Below two print statements should give the same results but they don't:

import re

s = '[a]'

print(*re.findall(r'\[.*]',s))
[a]

print(*re.findall(r'\[(.*)]',s))
a

3.6 is in security mode and not in bugfix mode.

Please, use 3.7.

Python 3.7.2 (default, Jan 16 2019, 19:49:22) 
[GCC 8.2.1 20181215 (Red Hat 8.2.1-6)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import re
>>> s = '[a]'
>>> print(*re.findall(r'\[.*]', s))
[a]
>>> print(*re.findall(r'\[.*]', s))
[a]
>>>

Fabrod, this has nothing to do with the square brackets. It is your assumption that the return value should be the same that is wrong. Cf.:

>>> re.findall(r'a.+', 'abc')
['abc']
>>> re.findall(r'a(.+)', 'abc')
['bc']

This is the correct documented behaviour.