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How to make regex for 3 slashes?

I want to write regex for following
students/ad34567-06c1-498c-9b15-cdbac695c1f2/data/sessions
Where students, data and sessions should be exact match.
i have tried this
[students]\[a-z]\[a-z]\[a-z]

You can try this regex, although your question is not clear to me.
^students\/([\w\-\d]+)\/data\/sessions$
Check here https://regex101.com/r/xnxwCX/1
you can grab the data in between students/, /data/session.

In your regex [students]\\[a-z]\\[a-z]\\[a-z] you are trying to match with word students in a character class [students] which will match one of the specified characters instead of matching the whole word.
To match a forward slash you have to use \/ instead of //. [a-z] is specified without a quantifier and will match 1 character from a-z.
To match your example string you might use
^students\/[a-z0-9]+(?:-[a-z0-9]+)+\/data\/sessions$
Regex demo
This part [a-z0-9]+(?:-[a-z0-9]+)+ matches one or more times a lowercase character or a digit [a-z0-9]+
Following a non capturing group repeated one or more times that will match a hyphen followed by matching one or more times a lowercase character or a digit (?:-[a-z0-9]+)+
You might also use [a-f0-9] if your characters are a -f

RegEx matching help: won't match on each appearence

I need to write a little RegEx matcher which will match any occurrence of strings in the form of
[a-zA-Z]+(_[a-zA-Z0-9]+)?
If I use the regex above it does match the sections needed but would also match onto the abc part of 4_abc which is not intended. I tried to exclude it with:
(?:[^a-zA-Z0-9_]|^)([a-zA-Z]+(_[a-zA-Z0-9]+)?)(?:[^a-zA-Z0-9_]|$)
The problem is that the 'not' matches at the beginning and end are not really working like I hoped they would. If I use them on the example
a_d Dd_da 4_d d_4
they would block matching the second Dd_da because the space was used in the first match.Sadly I can't use lookarounds because I am using JS.
So the input:
a_d Dd_da 4_d d_4
should match: a_d, Dd_da and d_4
but matches: a_d (there is a space at the end)
Is there another way to match the needed sections, or to not consume the 'anchor' matches?
I really appreciate your help.

You can make use of \b:
\b[a-zA-Z]+(_[a-zA-Z0-9]+)?\b
\b matches the (zero-width) point where either the preceding character or following character is a letter, digit or underscore, but not both. It also matches with the start/end of the string if the first/last character is a letter, digit or underscore.

Regex - Allow capital letters just at the beginning of words

I have to check for capital letters to exist just at the beginning of words.
My regex now looks like this:
/^([A-ZÁÉÚŐÓÜÖÍ]([a-záéúőóüöí]*\s?))+$/
It's at the words beginning works good, but if the problem not at the beginning of the word it's fails.
For example: John JohnJ got validated.
What should i alternate in my regex to works well?

In your regex pattern the space is optional, allowing combinations like JJohn or JohnJ - the key is to make it required between words. There are two ways to do this:
Roll out your pattern:
/^[A-ZÁÉÚŐÓÜÖÍ][a-záéúőóüöí]*(?:\s[A-ZÁÉÚŐÓÜÖÍ][a-záéúőóüöí]*)*$/
Or make the space in your pattern required, but alternatively allow it to be the end of line (this allows a trailing space though).
/^(?:[A-ZÁÉÚŐÓÜÖÍ][a-záéúőóüöí]*(?:\s|$))+$/
In both patterns I have removed some superfluous groups of your original and turned all groups into non-capturing ones.

You can do this: /^([A-ZÁÉÚŐÓÜÖÍ]{0,1}([a-záéúőóüöí]*\s?))+$/
With {a,b}, a is the least amount of characters it will match, whereas b is the most amount of characters it will match.
If there is ALWAYS going to be a capital letter at the beginning, instead you can simply use: /^([A-ZÁÉÚŐÓÜÖÍ]{1}([a-záéúőóüöí]*\s?))+$/
In this preceding case, {c}, c is the exact number of characters it will match.
Here is a resource with good information.

RegExp to match hashtag at the begining of the string or after a space

I have looked through previous questions and answers, however they do not solve the following:
https://stackoverflow.com/questions/ask#notHashTag
The closest I got to is this: (^#|(?:\s)#)(\w+), which finds the hashtag in half the necessary cases and also includes the leading space in the returned text. Here are all the cases that need to be matched:
#hashtag
a #hashtag
a #hashtag world
cool.#hashtag
##hashtag, but only until the comma and starting at second hash
#hashtag#hashtag two separate matches
And these should be skipped:
https://stackoverflow.com/questions/ask#notHashTag
Word#notHashTag
#ab is too short to be a hashtag, 3 characters minimum

This should work for everything but #hashtag#duplicates, and because JS doesn't support lookbehind, that's probably not possible to match that by itself.
\B#\w{3,}
\B is designed to match only between two word characters or two non-word characters. Since # is a non-word character, this forces the match to be preceded by a space or punctuation, or the beginning of the string.

Try this regex:
(?:^|[\s.])(#+\w{3,})(#+\w{3,})?
Online Demo: http://regex101.com/r/kG1nD5

/(\S)\1(\1)+/g matching all occurrences of three equal non-whitespace characters following each other

Its given: /(\S)\1(\1)+/g matches all occurrences of three equal non-whitespace characters following each other.
I don't understand why there is () around (\S) and 2nd (\1), but not around 1st (\1). Can anyone help in explaining how above regex works?
src: http://www.javascriptkit.com/javatutors/redev2.shtml
Thnx in advance.

The \S needs parentheses to capture its value, so you can refer back to the captured value with \1. \1 means "match the same text which capturing group #1 matched".
I believe there is a problem with this regex. You said you want to match "three equal non-whitespace characters". But the + will make this match 3 or more equal, consecutive non-whitespace characters.
The g on the end means "apply this regex over the entire input string, or globally".

The second set of parentheses is not necessary. It needlessly captures the repeated character a second time, while matching the same strings as this regex:
/(\S)\1\1+/g
Also, as #AlexD pointed out, the description should say that it matches at least three characters. If you replaced that regex with BONK in the string fooxxxxxxbar:
'fooxxxxxxbar'.replace(/(\S)\1\1+/g, 'BONK')
..you might expect the result to be fooBONKBONKbar from their description, because there are two sets of three 'x's. But in fact the result would be fooBONKbar; the first \1 matches the second 'x', and the \1+ matches the third 'x' and any 'x's that follow it. If they wanted to match just three characters, they should have left the + off.
I noticed several other sloppy descriptions like that, plus at least one outright error: \B is equivalent to (?!\b) (a position that's not a word boundary), not [^\b] (a character that's not a backspace). For that matter, their description of word boundaries--"the position between a word and a space"--is wrong, too. A word boundary isn't defined by any particular character, like a space--in fact, it can just as well be the absence of any character that creates one. The string:
Word
...starts with a word boundary because 'W' is a word character and, being first, it's not preceded by another word character. Similarly, the 'd' is not followed by another word character, so the end of the string is also a word boundary.
Also, a regex doesn't know from words, only word characters. The definition of a word character can vary depending on the regex flavor and Unicode or locale settings, but it always includes [A-Za-z0-9_] (ASCII letters and digits plus the underscore). A word boundary is simply a position that's between one of those characters and any other character (or no other character, as I explained earlier).
If you want to learn about regexes, I suggest you forget that site and start here instead: regular-expressions.info.