Well hi there!

Do you know the probability of selecting an even number out of 1 to 8? Obviously it is

p(even number out of 1-8) = ...

Now imagine you already selected an even number. The papers are not replaced, so there are 7 papers left. There are still 4 of them labeled with an uneven number. In addition there are 3 even numbers left, because you have already selected an even number in the first draw. So it is

p(even number out of 7 remaining papers) = 3/7.

p(both selected numbers are even) = p(even number out of 1-8) * p(even number out of 7 remaining papers)

Do you know what p(even number out of 1-8) is equal to?