The length of the cycles mod 10^k is simply Euler's phi function of 5^k: 5^(k-1) * 4 (or a factor of phi(5^k); AFAIK it is always exactly phi(5^k), although I don't have a proof of this handy).

The length of the even subset grows roughly as 2.5^k * 1.6. To see why, consider that the length of the cycle grows by a factor of 5 when incrementing k. Each all-even-digit power mod 10^k leads to 5 numbers mod 10^{k+1} which all share the same last k digits - i.e. their last k digits are even. We can model the k+1'th digit as being random, in which case we expect half of all those new numbers to consist entirely of even digits (one new digit, which is either odd or even, and k digits from the previous round that are all even). Thus, when incrementing k, the number of all-even-digit powers in the cycle will grow by approximately a factor of 2.5.