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Match the alphanumeric words(NOT NUMERIC-ONLY words) which have unique digits

Using regular expression, I want to select only the words which:
are alphanumeric
do not contain only numbers
do not contain only alphabets
have unique numbers(1 or more)
I am not really good with the regex but so far, I have tried [^\d\s]*(\d+)(?!.*\1) which takes me nowhere close to the desired output :(
Here are the input strings:
I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3
Expected Matches:
abc123
ab12s
hel1lo2haha3hoho4

You can use
\b(?=\d*[a-z])(?=[a-z]*\d)(?:[a-z]|(\d)(?!\w*\1))+\b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with \b, then:
(?=\d*[a-z]) - Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*\d) - Lookahead for a digit somewhere in the word
(?:[a-z]|(\d)(?!\w*\1))+ Repeatedly match either:
[a-z] - Any alphabetical character, or
(\d)(?!\w*\1) - A digit which does not occur again in the same word

Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/\b(?=[a-z]*\d)(?=\d*[a-z])(?!\w*(\d)\w*\1)[a-z\d]+\b/ig
RegEx Demo
RegEx Details:
\b: Word boundary
(?=[a-z]*\d): Make sure we have at least a digit
(?=\d*[a-z]): Make sure we have at least a letter
(?!\w*(\d)\w*\1): Make sure digits are not repeated anywhere in the word
[a-z\d]+: Match 1+ alphanumericals
\b: Word boundary

You could assert all the conditions using one negative lookahead:
\b(?![a-z]+\b|\d+\b|\w*(\d)\w*\1)[a-z\d]+\b
See live demo here
The important parts are starting match from \b and immediately looking for the conditions:
[a-z]+\b Only alphabetic
\d+\b Only numeric
\w*(\d)\w*\1 Has a repeating digit

You can use this
\b(?!\w*(\d)\w*\1)(?=(?:[a-z]+\d+)|(?:\d+[a-z]+))[a-z0-9]+\b
\b - Word boundary.
(?!\w*(\d)\w*\1) - Condition to check unique digits.
(?=(?:[a-z]+\d+)|(?:\d+[a-z]+)) - Condition to check alphanumeric words.
[a-z0-9]+ - Matches a to z and 0 to 9
Demo

Regex to allow alphanumeric, spaces, some special characters

I have this ^[a-zA-Z0-9 #&$]*$, but not working for me in few cases.
If someone types
A string that only consists of digits (e.g. 1234567)
A string starting with a special character (e.g. &123abc)
need to be rejected. Note that a special char can be in the middle and at the end.

You seem to need to avoid matching strings that only consist of digits and make sure the strings start with an alphanumeric. I assume you also need to be able to match empty strings (the original regex matches empty strings).
That is why I suggest
^(?!\d+$)(?:[a-zA-Z0-9][a-zA-Z0-9 #&$]*)?$
See the regex demo
Details
^ - start of string
(?!\d+$) - the negative lookahead that fails the match if a string is numeric only
(?:[a-zA-Z0-9][a-zA-Z0-9 #&$]*)? - an optional sequence of:
[a-zA-Z0-9] - a digit or a letter
[a-zA-Z0-9 #&$]* - 0+ digits, letters, spaces, #, & or $ chars
$ - end of string.

you can do it with the following regex
^(?!\d+$)\w+\S+
check the demo here

REGEX: \d not working

I have a text:
Wheels – F/R_ Schwalbe TABLE TOP/Schwalbe Black Jack 26x2.2
And regex to parse wheels size from that string:
/.*Wheels.*(\d*)x/
But it does not work. Besides, when i'm removing asterisk from regex, i'm getting number 6 as group match.

You need to make your .* before the digits lazy instead of greedy:
/.*Wheels.*?(\d*)x/
The .* will greedily consume everything up to the x, leaving nothing for the following \d*. Since * can validly match zero characters, an empty match for \d* is not an incorrect result.
By adding a ? to make a lazy .*? expression, it will try match as few characters as possible, allowing the following \d* to match all the numbers before the x.

You need to make your regex non-greedy because .* will consume your digits and \d* mean zero or no match
.*Wheels.*?(\d*)x
.*? mean match as many characters as few time as possible to stop .* to consume your digits
Follow this Demo for example
Alternately you can make it more efficient if there are no digits after Wheel and your desired values with following regex
.*Wheels[^\d]*(\d*)x
where [^\d]* mean matches anything except digits

Regex for digits and a plus sign

The conditions of the regex are as follows:
Starts with either digits or a '+' sign and ends with digits.
This is going to be used to validate a certain type of number. What I got so far is:
/^\d*|\+\d*$/
This regex seems to match any string though. How would a regex that matches my conditions look like?
The regex will be used in a JavaScript function.

I think you want something like this,
^(?:[+\d].*\d|\d)$
^ Asserts that we are at the start.
[+\d] Matches a plus symbol or a digit.
.* Matches any character zero or more times.
\d Matches a digit.
| OR
\d A single digit.
$ Asserts that we are at the end.
Use this if you want to match also a line which has a single plus or digit.
^[+\d](?:.*\d)?$
DEMO

You need to use anchors ^ and $ on both sides of your regex and make first part + or digit) optional.
You can use this regex:
^([+\d].*)?\d$
RegEx Demo

Javascript regular expression

I am making form and there is only one more thing which I cant figure it out :(
I need regular expression for password which must be at least 7 characters long. There can be small and big letters and must contain at least one number.
I tried
[0-9]+[a-zA-Z]){7}$

You can use lookahead:
^(?=.*\d)[a-zA-Z\d]{7,}$
(?=.*\d) is a lookahead which checks for a digit in the string. Basically, .* matches the whole string and then backtracks 1 by 1 to match a digit. If it matches a digit, the regex engine comes back to its position before match. So, it just checks for a pattern.
{7,} is a quantifier which matches previous pattern 7 to many times
^ is the beginning of a string