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HTML+CSS:Is there a way to have a container(modal) in "vacuum" from the rest of the page so no other styling applied to it?

I have downloaded a free bootstrap template which has a lot of pages with a lot of styles and scripts.
When I try to add an html container to that page(which is a modal/popup), all the styles from body to headers apply to it which breaks it completely.
So because I don't want to create a class/id for all and some width/heights on the parent and body divs styling are impossible to avoid without breaking the template flow, I am wondering if there is a way to create a html container with some option that if you add it, absolutely no other styling applies to it and I can style it as I wish irrespective of what happens around it?
Could z-index work here?
L.EDIT
I have added the code here on codepen [codepen modal][1]enter link description here
The modal should look like in the codepen but instead it is spread out like it is here in this screenshot.

Your styling for your custom element must be of higher specificity than the other styling that is declared. Take a look at this great article by Emma Bostian :
https://dev.to/emmabostian/css-specificity-1kca#:~:text=CSS%20Specificity%20is%20the%20set,present%20on%20the%20element's%20style.

There's a following property in CSS:
#Element {
all: initial;
}
This should reset all the styling from the parent elements including the container in which the element is placed and the body.
Try this and declare the styles after.

How to make an element's CSS NOT inherit from others?

I want the element to use only css that are in the "A" section and ignore the "B" section.
Is it possible?If javascript can do this, How?
Thanks you.

You can not do that with the example you've provided. The C in CSS stands for Cascading, the styling rules cascade down the DOM tree.
You have to reset the styling of the element to what you want with a more specific selector, e.g. #Examplewrapper input{}. By using a more specific selector, it'll overwrite/suplement the previous styling, without the need for !important.
Alternatively, you can set the most upper selector more specific, e.g. #content input{}. This way, when you place a form in the #footer, it will not have the styling, as #content doesn't have a #footer in it (it cant cascade).
I do recommend to define a general input as you have. This way, all forms have the same font, size and styling throughout your website. If you want another color border, you only have to change that one settings. This is the way many (profesional) sites work, because it is the most efficient.

This is how the inheritance works. You can only overwrite styles if others are set globally (i.e. for all input elements).
You can always limit the global styles of input with some classname, like input.myStyle so the raw input will have no styles set.

Moving an element into a new div - doesn't take on its new css properties

So, I have an element that has some "pre-existing" behavior attached to it. So, I found that just moving it (as required by some new requirements) retains the existing behaviors = good. But the issue is, when I move the element I need to give it "new styles".
So, if I have something like this:
<div id="existingStructure">
<div id="legacyElement"></div>
</div>
Now, that has pre-existing styles attached to both. I can rearrange these styles etc.. but I can't change them. The styles are attached to the "id's" rather than a class definition. I believe I can change that if needed.
Now, I need to move "legacyElement" when certain things happen to a "new div".
I just
jQuery('#newStructure').append('#legacyElement');
<div id="newStructure">
<div id="legacyElement"></div>
</div>
Unfortunetly, the styles I have on newStructure don't seem to be applying to *legacyElement" when it gets moved here dynamically.
I was thinking of moving all the styles to a class rather than associated to the ids, and when I move it.. I just jQuery().addClass / jQuery().removeClass etc...
Is there a better/easier more robust way that I can just have the legacyElement loose its styles when it sits under existingStructure and then get new ones when moved to "newStructure" etc.. and vice versa. That element "legacyElement" will be pinging back and forth.. so, I need it to have the styles under each parent div as it goes there.
so when an action happens on page, I move it back:
jQuery('#existingStructure').append('#legacyElement');
If I am not succinct enough, please let me know.
The EXISTING styles are in an external CSS file and are like so..
#existingStructure {
// bunch of css
}
#existingStructure .item1 input[type="text"] {
// bunch of css
}
#legacyElement{
// bunch of css
}
and new styles are sorta the same except 'additional styles' might be applied.
#newStructure {
// bunch of css
}
#newStructure .item1 input[type="text"] {
// bunch of css
}

You can certainly target your div styles by their parents:
#existingStructure #legacyElement {some styles}
#newStructure #legacyElement {some other styles}
To explain futher, this arrangement should result in greater specificity, overriding styles that are simply applied to either #existingStructure or #legacyElement. I'm hoping no one did anything foolish like using !important on them.

Short answer: It should.
Here's an example I quickly made in jsFiddle: http://jsfiddle.net/CCm4J/1/
So then why isn't yours? Most likely you have css rules that are embedded that apply only when in the existingStructure id/class perhaps? Without see more of your css I'm not sure how specific I can get. I would just verify that your css rules are allowed to apply outside of existingStructure (and even that existingStructure might have rules for its parent too!)

Change CSS properties of non-existing elements

I want to change the dimensions of a set of elements that are not created yet. They will be created by a JavaScript script that I don't have access to. JQuery's css() function would apply the changes only on existing items, while I want the code to work like if I had set CSS properties in a CSS file.
Can anyone help me do it?

One option would be to dynamically create a style element:
$("<style type='text/css'> .redbold{ color:#f00; font-weight:bold;} </style>").appendTo("head")
$("<div/>").addClass("redbold").appendTo("body");
Taken from here: Create a CSS rule / class with jQuery at runtime.
Here's a possible IE alternative (see Creating a CSS class in jQuery):
document.styleSheets[0].addRule('body', 'background: green', -1);
For reference, see the documentation for addRule.

Remove/change Internal styles sheet styles with JavaScript/jQuery

I want to be able to remove/change the internal style sheets values through JavasScript. It has to be through JS because I cannot edit the html since I am using an application that does not allow me to, but I can use JS. How can I change the value for the background or remove it completely? Or is there another way to accomplish this?
<body>
<head>
<style type="text/css">
#company-header{background:#000 !important;}
</style>
</head>
<div id="company-header"></div>
</body>

If your just going to change small bits of css, using jQuery's css() is your best option, but it does not always recognize !important, for that you would probably need cssText, like this:
$('#id').css('cssText', 'height: 200px !important');
This replaces all the css, so everything needs to be defined in the new css rules that are added.
If you are changing a lot of css, or just want to make it easier for the next time, you could remove all inline css and add an external stylesheet instead.
To remove all inline styles you would do something like this:
$(document).removeAttr('style');
or for div's only
$('div').removeAttr('style');
Depending on how many styles there are, this could take som time to process.
Then to add a new stylesheet do:
(function() {
var myCSS = document.createElement('link');
myCSS.rel = 'stylesheet';
myCSS.type = 'text/css';
myCSS.src = '/styles/mystylesheet.css';
var place = document.getElementsByTagName('script')[0];
place.parentNode.insertBefore(myCSS, place);
})();
Edit:
function removeCSS() {
var badCSS = document.getElementsByTagName('style')[0];
$(badCSS).remove();
});
This will remove all the markup in the internal stylesheet, but since the styles are already loaded it will make absolutely no difference.
Internal styles will always override external styles, but for one exeption, if the external style is !important. If both the external and internal styles are !important, the internal style will be used.
Loading an external stylesheet dynamicly with javascript will only work if everything you are trying to override is set to !important in the external stylesheet, and not set to !important in the internal stylesheet.
Changing the styles directly in the DOM with jQuery's css() and using the cssText option to override the !important set in the internal stylesheet may be your only viable option if there is absolutely no way to alter the html file and remove the internal stylesheet.

EDIT: OK, now that we understand that this question is really just about how to override the !important style declaration when setting styles via javascript, this is a duplicate of this posting: Overriding !important style.
The two possible answers there are to set a whole style string on the object or to create a new stylesheet that refers to this object.
Previous answer before question was edited:
You can just set some style directly on the object if you want like this. This will override anything that comes from a style sheet so you don't have to mess with the style sheet.
$("#company-header").css("background-color", "#FFFFFF");
or
$("#company-header").css("background", "none");
A more extensible way of modifying the look of an object or groups of objects is to based the style settings on class names and then add/remove class names using jQuery. This has the effect of switching which style sheet rules apply to a given object without having to directly manipulate the style sheets themselves:
$("#company-header").removeClass("oldClass").addClass("newClass");

$(document).ready(function(){
$('style').remove();
});
This code will remove the all internal css from the site. I used this but style tag will be visible in the page source.