(ii) To find the probability distribution of \(X\), count the occurrences of each sum in the table and divide by the total number of outcomes (36). The probabilities are:

(iii) The expected value \(E(X)\) is calculated as:

\(E(X) = \sum p_i x_i = 2 \times \frac{1}{36} + 4 \times \frac{2}{36} + 6 \times \frac{5}{36} + 7 \times \frac{4}{36} + 8 \times \frac{4}{36} + 9 \times \frac{4}{36} + 10 \times \frac{4}{36} + 11 \times \frac{8}{36} + 12 \times \frac{4}{36}\)

\(= \frac{312}{36} = \frac{26}{3} \approx 8.67\)

(iv) The probability that \(X\) is greater than \(E(X)\) is:

\(P(X > 8.67) = P(X = 9, 10, 11, 12) = \frac{4}{36} + \frac{4}{36} + \frac{8}{36} + \frac{4}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.556\)