Finding the last two decimal places.....

Finding the last two decimal digits of this number, we know before that the order of [3]mod43 is 42. As a result i was thinking it may be linked in with this with applying the fact that [3]^42 mod 43=1mod 43 , then simplifying the (9^9) from this fact, on the other hand it may not have anything to do with this, The question is designed to work it out systematically via general results instead of just throwing it into a computer programme to read the last two decimals off,

many thanks in advance.

I have a additional question in finding the order of [a] mod n, its the smallest positive integer such that [a]^t mod n=1mod n. Hence just say if we had [2]mod 75, we know that [2]^40mod75=1mod75, by finding phi(75), im just a bit confused from here, do we check all factors of 40, to see whether they are equal to 1mod75, if so the lowest of these will be our order. I know that 2^20mod75=1mod75, then from this do we check all factors of 20, say 10,2,4,5?