Taking into account your last comment to your Q,
>>> x = [['asgy200;f','ssll100','time is: 10h:00m:12s','xxxxxxx','***','','asgy200;f','frl5100','time is: 00h:00m:05s','ooo']]
>>> print all([w[-3:-1]!='00' for r in x for w in r if w.startswith('time is: ')])
True
>>>
all and any are two useful builtins...
The thing operates like this, the slower loop is on the sublists (rows) of x, the fastest loop on the items (words)in each row, we pick up only the words that startswith a specific string, and our iterable is made of booleans where we have true if the 3rd last and 2nd last character of the picked word are different from'00'. Finally the all consumes the iterable and returns True if all the second fields are different from '00'.
HTH,
Addendum
Do we want to break out early?
all_secs_differ_from_0 = True
for row in x:
for word in row:
if word.startswith('time is: ') and word[-3:-1] == '00':
all_secs_differ_from_0 = False
break
if not all_secs_differ_from_0: break