If the two triangles are similar then their angles have the same measure. This also implies that the quotient between a side of one triangle and its corresponding side in the other one is the same for the three pairs of sides. The sides of the large triangle are 9, 2+y and 12 and their corresponding sides in the small triangle are 3, 2 and x. Then since the quotient between corresponding sides is always the same we get:

So for x we get:

We multiply both sides by x and we get:

And we divide both sides by 3:

Then for y we get:

We can multiply both sides by 2:

And we substract 2 from both sides:

So x=4 and y=4. Then the answer to part 1 is option A and the answer to part 2 is option B.