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What Regex would capture both the beginning and end from of a string?

I am trying to edit a DateTime string in typescript file.
The string in question is 02T13:18:43.000Z.
I want to trim the first three characters including the letter T from the beginning of a string AND also all 5 characters from the end of the string, that is Z000., including the dot character. Essentialy I want the result to look like this: 13:18:43.
From what I found the following pattern (^(.*?)T) can accomplish only the first part of the trim I require, that leaves the initial result like this: 13:18:43.000Z.
What kind of Regex pattern must I use to include the second part of the trim I have mentioned? I have tried to include the following block in the same pattern (Z000.)$ but of course it failed.
Thanks.
Any help would be appreciated.

There is no need to use regular expression in order to achieve that. You can simply use:
let value = '02T13:18:43.000Z';
let newValue = value.slice(3, -5);
console.log(newValue);
it will return 13:18:43, assumming that your string will always have the same pattern. According to the documentation slice method will substring from beginIndex to endIndex. endIndex is optional.

as I see you only need regex solution so does this pattern work?
(\d{2}:)+\d{2} or simply \d{2}:\d{2}:\d{2}
it searches much times for digit-digit-doubleDot combos and digit-digit-doubleDot at the end
the only disadvange is that it doesn't check whether say there are no minutes>59 and etc.
The main reason why I didn't include checking just because I kept in mind that you get your dates from sources where data that are stored are already valid, ex. database.

Solution
This should suffice to remove both the prefix from beginning to T and postfix from . to end:
/^.*T|\..*$/g
console.log(new Date().toISOString().replace(/^.*T|\..*$/g, ''))
See the visualization on debuggex
Explanation
The section ^.*T removes all characters up to and including the last encountered T in the string.
The section \..*$ removes all characters from the first encountered . to the end of the string.
The | in between coupled with the global g flag allows the regular expression to match both sections in the string, allowing .replace(..., '') to trim both simultaneously.

Get all the WORDS except one specific word

I want to get all the words, except one, from a string using JS regex match function. For example, for a string testhello123worldtestWTF, excluding the word test, the result would be helloworldWTF.
I realize that I have to do it using look-ahead functions, but I can't figiure out how exactly. I came up with the following regex (?!test)[a-zA-Z]+(?=.*test), however, it work only partially.
http://refiddle.com/refiddles/59511c2075622d324c090000

IMHO, I would try to replace the incriminated word with an empty string, no?

Lookarounds seem to be an overkill for it, you can just replace the test with nothing:
var str = 'testhello123worldtestWTF';
var res = str.replace(/test/g, '');

Plugging this into your refiddle produces the results you're looking for:
/(test)/g
It matches all occurrences of the word "test" without picking up unwanted words/letters. You can set this to whatever variable you need to hold these.

WORDS OF CAUTION
Seeing that you have no set delimiters in your inputted string, I must say that you cannot reliably exclude a specific word - to a certain extent.
For example, if you want to exclude test, this might create a problem if the input was protester or rotatestreet. You don't have clear demarcations of what a word is, thus leading you to exclude test when you might not have meant to.
On the other hand, if you just want to ignore the string test regardless, just replace test with an empty string and you are good to go.

Regular expression (match function), javascript

I think this is a very basic question, but I really can't understand the concept. I have the following regular expression:
var t = '11:59 am';
t.match(/^(\d+)/);
Now, according to my understanding when I print the value I should just get 11 since I am just checking for digits. However, I get 11,11. I have to use 0th element to pick the required value like t.match(/^(\d+)/)[0].

This is because you are using a capture group, (), around the digits. Try replacing this with:
t.match(/^\d+/);
Note: this will still return an array, because that's just what .match() does.

match() always returns an array if there are any matches. Element [0] is the whole match, and element [1] is what is inside the first set of parentheses.

What's wrong with this regular expression to find URLs?

I'm working on a JavaScript to extract a URL from a Google search URL, like so:
http://www.google.com/search?client=safari&rls=en&q=thisisthepartiwanttofind.org&ie=UTF-8&oe=UTF-8
Right now, my code looks like this:
var checkForURL = /[\w\d](.org)/i;
var findTheURL = checkForURL.exec(theURL);
I've ran this through a couple regex testers and it seems to work, but in practice the string I get returned looks like this:
thisisthepartiwanttofind.org,.org
So where's that trailing ,.org coming from?
I know my pattern isn't super robust but please don't suggest better patterns to use. I'd really just like advice on what in particular I did wrong with this one. Thanks!

Remove the parentheses in the regex if you do not process the .org (unlikely since it is a literal). As per #Mark comment, add a + to match one or more characters of the class [\w\d]. Also, I would escape the dot:
var checkForURL = /[\w\d]+\.org/i;

What you're actually getting is an array of 2 results, the first being the whole match, the second - the group you defined by using parens (.org).
Compare with:
/([\w\d]+)\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl"]
/[\w\d]+\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org"]
/([\w\d]+)(\.org)/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl", ".org"]
The result of an .exec of a JS regex is an Array of strings, the first being the whole match and the subsequent representing groups that you defined by using parens. If there are no parens in the regex, there will only be one element in this array - the whole match.

You should escape .(DOT) in (.org) regex group or it matches any character. So your regex would become:
/[\w\d]+(\.org)/
To match the url in your example you can use something like this:
https?://([0-9a-zA-Z_.?=&\-]+/?)+
or something more accurate like this (you should choose the right regex according to your needs):
^https?://([0-9a-zA-Z_\-]+\.)+(com|org|net|WhatEverYouWant)(/[0-9a-zA-Z_\-?=&.]+)$

Using JavaScript RegExp to match the first item in a list surrounded by parentheses

I have a string of text I'm trying to parse with RegExp in JavaScript. Let's say it looks like this:
var myString = "This is a string of text (item item item) and more text here.";
I need to match the first occurrence of the word 'item' based on the fact that it is the first item inside the parentheses. I can't figure out how to write a pattern that will match ONLY the first item inside a set of parentheses.
Some of you may want to think of it like this: Pretend I'm parsing a string of Lisp and want to match all cars.
Thanks in advance.

If you can avoid checking for an opening parenthesis you could use a look ahead for the closing parenthesis.
item(?=[^\)]*\))
You can also use capturing groups with:
\(.*?(item).*?\)
EDIT
For a word that is repeated somewhere in the parentheses at least once:
\(.*?(\b\w+\b).*?\1.*?\)
EDIT 2
For just the first alphanumeric set of characters in a parentheses:
\([^\w]*(\b\w+\b).*?\)
Or a simpler alternative:
\(.*?\b(\w+)\b.*?\)

Use something like this:
/\((\w*?)(?:\s|\))/
Live example: http://tinkerbin.com/EJyjEX30 (click run to run the code))

You cannot make a pattern that doesn't match the opening parenthesis, since Javascript regexes don't support lookback.
You probably need to update your lexer code to understand capturing groups (or general functions or something convenient...) or you need to restructure you code so that it can deal with that extra paren you can't get rid of.