How to print the file name only when a searched file content pattern is found

I've a Linux search command working for a specific pattern within a block in multiple files.
But I'm missing the file name in the output for the files where the specific pattern was found. Since I could have 1000+ files this is a big problem for me, and I can't have all the file names that is checked printed, only the once where the pattern exists.

Update
Got some feedback. I have changed the id:s for the start and end tags within the block. Good to Know if you think some (old) answers look strange.

• 4 => # unique_start_tag
• 7 => # unique_end_tag

These files are created from a template so the tags Will always be there!

$ cat file1.tr
1
2
ID   : file1
3
# unique_start_tag
55
66
# unique_end_tag
8
9

$ cat file2.tr
1
2
ID   : file2
5
3
# unique_start_tag
5
6
# unique_end_tag
8
9

$ cat file3.tr
1
2
ID   : file3
5
3
# unique_start_tag
5
kalle
6
# unique_end_tag
8
9

What I tried (found after googling...):

1)

$ find ./ -type f -name "*.tr" | xargs sed -n '/# unique_start_tag/!b;:a;/# unique_end_tag/!{$!{N;ba}};{/55/p}'
# unique_start_tag
55
66
# unique_end_tag

2)

    $ find ./ -type f -name "*.tr" | xargs sed -n '/# unique_start_tag/!b;:a;/# unique_end_tag/!{$!{N;ba}};{/5/p}'
# unique_start_tag
55
66
# unique_end_tag
# unique_start_tag
5
6
# unique_end_tag
# unique_start_tag
5
kalle
6
# unique_end_tag

The "wildcard approach" in the output is okay. But since the file name is missing in the printout, it's pretty useless since it can be 1000+ files.

Question

a) How can the commands in "2" above be changed to also print the filename prior to the file content for files when there is a hit? (Extra line brake would be nice as well in-between each file output result)

So, what I would wish from "2" above:

file1.tr
# unique_start_tag
55
66
# unique_end_tag

file2.tr
# unique_start_tag
5
6
# unique_end_tag

file3.tr
# unique_start_tag
5
kalle
6
# unique_end_tag

b) Possible alternative to a? The files I'm dealing with have an "ID" line within the file that could be used to parse the "filename".
"ID : file1" is the line for file1. So, it is okay to parse that line instead as an alternative solution to "a". Since this ID maps against the real filename. How can it be solved, if you think it's easier than "a"? Example:

   $ find ./ -type f -name "*.tr" | xargs sed -n '/ID/p'
ID   : file1
ID   : file2
ID   : file3

But the general requirements is of course the same.

c) How can the printout be limited to only contain the specific "pattern line"? Only "5" respective "55" lines in the examples. To compare with the output seen in 1 and 2 above, where the complete block is printed!

d) How can the command be changed to exclude the start and end tags from the output?

All provided solution must work in Git bash terminal. And one-liners is good.
$ git version
git version 2.35.1.windows.2

Will this awk version do?

find . -type f -name \*.tr | xargs -n 1 awk 'NR==1{print FILENAME} /4/,/7/{print} END{print ""}' 
./file1.tr
4
55
66
7

./file3.tr
4
5
kalle
6
7

./file2.tr
4
5
6
7

Just grep it:

$ ls f[1-3]
f1  f2  f3

$ cat f[1-3]
a
b
a

$ grep -ril a f[1-3]
f1
f3

And if you want to parse files add them to an array and loop over it:

mapfile -t arr < <(grep -ril a f[1-3])

for i in "${arr[@]}"; do
    printf "$i "; grep a "$i"
done

f1 a
f3 a

This might work for you (GNU sed):

sed -n '/4/{:a;N;/7/!ba;/55/{F;G;p}}' file1 file2 file3 filen

Gather up the lines between two regexps and then test for another inclusive regexp. If the match is true, output the current filename (F), add an empty line (G) and print (p) the matched range.

If only the first match is required add the q command i.e

sed -n '/4/{:a;N;/7/!ba;/55/{F;G;p;q}}' file1 file2 file3 filen