Alternatively, expanding by Binomial Theorem, we have:

2^95
= (33-1)^19
= (33^19 – comb(19, 1)*33^18 +......- comb(19,17)*33^2 +  comb(19,18)*33 – 1)
= M(99) + 19*33 -1,
= M(99) +  18*33 + (33-1)
= M(99) + 32

Accordingly,

2^95 (mod 99) = 32, so that:
2^99 (mod 99) = (32*16) mod 99

But, 32*16
= (33-1)(15+1)
 = 33*15 +(33-15-1)
= M(99) + 17, since 15 is divisible by 3, so that 33*15 is divisible by 99

Accordingly, 
2^99 (mod 99) = 17

Consequently, the required remainder is 17.