Here's the very basic way to so it:
The totals on two dice and the different ways they can occur are:
2: 1, 1 one way
3: 2, 1; 1, 2 two ways
4: 3, 1; 2, 2, 1, 3 three ways
5: 4, 1; 3, 2, 2, 3, 1, 4 four ways
6: 5, 1; 4, 2, 3, 3, 2, 3, 1, 5 five ways
7: 6, 1; 5, 2; 4, 3, 3, 4, 2, 5, 1, 6 6 ways
8: 6, 2; 5, 3; 4, 4, 3, 5, 2, 6 five ways
9: 6, 3; 5, 4, 4, 5, 3, 6 four ways
10: 6, 4; 5, 5; 4, 6 three ways
11: 6, 5; 5, 6 two ways
12: 6, 6 one way
A total of ways.
The even numbers are 2: one way, 4: three ways, 6: five ways, 8: five ways, 10: three ways, 12: one way, a total of 1+ 3+ 5+ 5+ 3+ 1= 18 ways. Of those five give a total of 8: the probability that a pair of dice, given that the sum is even, have a sum of 8 is 5/18.
Once again, I didn't read all of the posts carefully- Plato has already suggested this and the OP did it properly.