2

5^n always ends in 5

4^n where n ends in 5 always ends in 24

3^n where n ends in 24 always ends in 81

2^n where n ends in 81 ends in 2

We're reducing 2^x mod 10, so by Euler's Totient we can reduce x mod phi(10)=4.
Since x = 3^y where y is even, we have x = 9^z for some integral z. Reducing the base mod 4 shows, x mod 4 = 1, so 2^x mod 10 = 2.

Edit: oops late to the punch

Edit: incorrect