(i) To find the probability of event R, we list the pairs where one score is exactly 3 greater than the other: \((1, 4), (2, 5), (3, 6), (4, 7), (5, 8)\). Each pair can occur in two ways (e.g., \((1, 4)\) and \((4, 1)\)). Thus, there are \(5 \times 2 = 10\) favorable outcomes. The total number of outcomes when the die is thrown twice is \(8 \times 8 = 64\). Therefore, \(P(R) = \frac{10}{64}\).

(ii) For event S, we need the product of the scores to be more than 19. The pairs satisfying this condition are \((3, 8), (3, 7), (4, 8), (4, 7), (4, 6), (4, 5), (5, 8), (5, 7), (5, 6), (6, 8), (6, 7), (7, 8)\). Additionally, \((5, 5), (6, 6), (7, 7), (8, 8)\) also satisfy the condition. The total number of favorable outcomes is \(24 + 4 = 28\). Thus, \(P(S) = \frac{28}{64}\).

(iii) To determine if R and S are independent, we check if \(P(R \cap S) = P(R) \times P(S)\). From the mark scheme, \(P(R \cap S) = \frac{4}{64}\). Calculating \(P(R) \times P(S) = \frac{10}{64} \times \frac{28}{64} = \frac{280}{4096}\), which simplifies to \(\frac{35}{512}\), not equal to \(\frac{4}{64}\). Therefore, events R and S are not independent.