Match regexp at the end of the string with AWK - awk

Question

I am trying to match two different Regexp to long strings with awk, removing the part of the string that matches in a 35 characters window.
The problem is that the same bunch of code works when I am looking for the first (which matches at the beginnng) whereas fails to match with the second one (end of string).
Input:
Regexp1(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)Regexp2
Desired output
(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)
So far I used this code that extracts correctly Regexp1, but, unfortunately, is not able to extract also Regexp2 since indexed of RSTART and RLENGTH for Regexp2 are incorrect.
Code for extracting Regexp1 (correct output):
awk -v F="Regexp1" '{if (match(substr($1,1,35),F)) print substr($1,RSTART,RLENGTH)}' file
Code for extracting Regexp2 (wrong output)
awk -v F="Regexp2" '{if (match(substr($1,length($1)-35,35),F)) print substr($1,RSTART,RLENGTH)}' file
Despite the indexes for Regexp1 are correct, for Regexp2 indexes are wrond (RSTART=13). I cannot figure out how to extract the second Regexp.

Answer 1

Considering that your actual Input_file is same as shown samples, if this is the case could you please try following then(good to have new version of awk since old versions may not support number of times logic for regex).
awk '
match($0,/\([0-9]+\){5}.*\([0-9]\){4}/){
print substr($0,RSTART,RLENGTH)
}' Input_file
In case your number of parenthesis values are not fixed then you could do like as follows:
awk '
match($0,/\([0-9]+\){1,}.*\([0-9]\){1,}/){
print substr($0,RSTART,RLENGTH)
}' Input_file

Answer 2

If this isn't all you need:
$ sed 's/Regexp1\(.*\)Regexp2/\1/' file
(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)
or using GNU awk for gensub():
$ awk '{print gensub(/Regexp1(.*)Regexp2/,"\\1",1)}' file
(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)
then edit your question to be far clearer with your requirements and example.