In the question here
Find the last two digits of the given number
The method for finding the last two digits of a single integer is fairly clear. However, when confronted with an expression like the one in this question, I am unsure of how to proceed. Ideally, I would like to avoid using any techniques asides from congruence.
Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.
So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2\equiv 8^2=64 \mod 100$
And $3^{305} = 3*(3^2)^{152} = 3*(10 - 1)^{152} = 3(...... + 152*10*(-1)^{151} + (-1)^{152})\equiv 3(-20 + 1) =-57\equiv 43\mod 100$.
And $11^{151} = (10 + 1)^{151} = ...... + {151\choose 2}10^2 + 151*10 + 1\equiv 11\mod 100$.
So $808^2 + 3^{305} \cdot 11^{151} \equiv 64 + 43 \cdot 11 = 64 + 473\equiv 37 \mod 100$.
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Oh, for goodness sake!
$3^2*11 = (10-1)(10+1) =100-1\equiv -1$
So $3^{2k}11^k =(99)^k \equiv (-1)^k\mod 100$
so
$3^{305}\cdot 11^{151}= 3^3*3^{302}\cdot 11^{151}=$
$27*(-1)^{151} \equiv -27\mod 100$.
So $808^2 + 3^{305}11^{151}\equiv 64-27= 37\mod 100$.