(i) To find \(P(X = 9)\), we consider the combinations of scores that sum to 9: (1, 4, 4), (2, 3, 4), and (3, 3, 3). The probabilities are calculated as follows:

\(P(1, 4, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 3 = \frac{3}{64}\)

\(P(2, 3, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 6 = \frac{6}{64}\)

\(P(3, 3, 3) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\)

Adding these probabilities gives \(P(X = 9) = \frac{3}{64} + \frac{6}{64} + \frac{1}{64} = \frac{10}{64}\).

(ii) The completed probability distribution table is:

(iii) To determine if \(R\) and \(S\) are independent, we calculate:

\(P(S) = \frac{6}{64}\)

\(P(R \cap S) = \frac{3}{64}\)

Since \(P(R \cap S) \neq P(R) \times P(S)\), events \(R\) and \(S\) are not independent.