We need a 3 or 6 on the first roll => (3,1)....(3,6) or (6,1) ... (6,6) => 12 of these.
Consider all the combinations that add up to 8:
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2).
However, (2, 6), (3, 5), and (6,2) were already in the previous event. So we do not need to double count. => 2 of these.
The total space is 6*6 = 36.
Therefore, (14+2)/36 = 16/36 = 4/9.
