(i) To find the probability of event A, we list the pairs where the scores differ by 3 or more: (1,4), (1,5), (1,6), (2,5), (2,6), (3,6) and their reverses: (4,1), (5,1), (6,1), (5,2), (6,2), (6,3). There are 12 such pairs out of 36 possible outcomes.

Thus, \(P(A) = \frac{12}{36} = \frac{1}{3}\).

(ii) For event B, we need the product of the scores to be greater than 8. The pairs are: (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6). There are 20 such pairs.

Thus, \(P(B) = \frac{20}{36} = \frac{5}{9}\).

(iii) Events A and B are not mutually exclusive because \(P(A \cap B) \neq 0\). Specifically, there are 6 outcomes that satisfy both conditions: (2,5), (2,6), (3,6), (5,2), (6,2), (6,3). Therefore, \(P(A \cap B) = \frac{6}{36}\), so they are not mutually exclusive.