If a is to be chosen at random from the integers...

For a+b to be even, either both a and b must be even or both a and b must be odd.

Since we will be choosing 5 possible "a" numbers and 5 possible "b" numbers, the total number of possibilities will be 25. The question is: How many of those will fit the above criteria?

We have 3 ways that "a" can be odd and 2 ways that "b" can be odd. So that is 6 possible ways (3x2).
We have 2 ways that "a" can be even and 3 ways that "b" can be even. So that is 6 possible ways.

So, all told, we have 12 possible successes out of 25 possible choices. So the probability will be 12/25.