Last two digits of $9^{1500}$

$(10-1)^{1500} = 10^{1500} - \binom{1500}{1}10^{1499}{1^1} + ... + \binom{1500}{1498}10^21^{1498} - \binom{1500}{1499}10^11^{1499} + 1^{1500}$ Only the last term is not divisible by hundred. Thus, $9^{1500} \equiv 1 \mod 100$ $i.e.$ the last two digits are 01.

Calculating the last two digits is like taking the number $\mod 100$.

$$9 \equiv 9\mod 100\\ 9^2 \equiv 81\mod 100\\ 9^3=729\equiv 29\mod 100\\ 9^4=9\cdot 9^3 \equiv 9\cdot 29 = 261\equiv 61\mod 100\\ 9^5\equiv 9\cdot 61 = 549\equiv 49\mod 100\\ 9^6 \equiv 9\cdot 49 = 441\equiv 41\mod 100\\ 9^7\equiv 9\cdot 41 = 369\equiv 69\mod 100\\ 9^8\equiv 9\cdot 69 = 721\equiv 21\mod 100\\ 9^9\equiv 9\cdot 21 = 189\equiv 89\mod 100\\ 9^{10}\equiv 9\cdot 89 = 801\equiv 1\mod 100\\ $$

Now, knowing that $9^{10}\equiv 1\mod 100$, your remainging task should be easier.