First of all, in these types of problems, note that you can always bring the modulo operator "inside".
a^b mod c=(a mod c)^b mod c (Standard formulae for modular arithmetic)
Thus 8^11 mod 65 = (8 mod 65)^11 mod 65=8^11 mod 65 !!! (not much help, huh)
Lets try something else: 8^11 mod 65 = ( 8 x (8^10) ) mod 65 [you can try any variety]
= ( 8 x (8 mod 65) ^10 ) mod 65 ) = 8 x (8 ^10) mod 65 = still no help
What if : 8^11 mod 65 = 8 x (8^2)^5 mod 65 = 8 x 64 ^ 5 mod 65 = ( 8 x (64 mod 65) ^ 5 ) mod 65 = ( 8 x (-1) ^ 5 ) mod 65 = (-8) mod 65 = (65+(-8)) mod 65 = 57
Obviously you could as well keep the answer as "-8", but when you say remainder is -8, it is as good as saying remainder is 57.
Also if you question how I could come up with that weird "8 x (8^2)^5 mod 65" thing, I would tell I was looking for some power of 8 that is close to 65, as close as possible. Why ? Because then I could insert the modulo operator inside which will help me in reducing the entire thing down a bit. Take for example, If I got some factor of 8 to be 67 which was the closest to 65 , I would try to express the LHS in term of "67 to the power something". Then I would bring the "modulo 65" inside the "67 to the power something". That will give me just "2 to the power something", which is much better.
