(a) The sum of probabilities must equal 1: \(p + q + 0.65 = 1\). Therefore, \(p + q = 0.35\).

Using the expected value: \(E(X) = 0 \times 0.6 + 1 \times p + 2 \times q + 3 \times 0.05 = 0.55\).

This simplifies to \(p + 2q + 0.15 = 0.55\), so \(p + 2q = 0.4\).

Solving the equations \(p + q = 0.35\) and \(p + 2q = 0.4\), we find \(q = 0.05\) and \(p = 0.3\).

(c) The probability that \(X = 1\) in at least 3 of 12 turns is \(1 - P(0, 1, 2)\).

\(P(0, 1, 2) = \binom{12}{0} (0.3)^0 (0.7)^{12} + \binom{12}{1} (0.3)^1 (0.7)^{11} + \binom{12}{2} (0.3)^2 (0.7)^{10}\).

Calculating these gives \(0.01384 + 0.07118 + 0.16779 = 0.25299\).

Thus, the probability is \(1 - 0.25299 = 0.747\).

(d) The probability that Jim first succeeds on the 9th turn is \((0.95)^8 \times 0.05\).

This evaluates to \(0.0332\).