(i) The probability that exactly one rabbit is white can be calculated by considering two cases: one white and one non-white rabbit.

Case 1: First rabbit is white, second is non-white: \(\frac{6}{9} \times \frac{3}{8}\).

Case 2: First rabbit is non-white, second is white: \(\frac{3}{9} \times \frac{6}{8}\).

Adding these probabilities gives: \(\frac{6}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{6}{8} = \frac{1}{2}\).

(ii) The probability distribution table is constructed as follows:

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

\(E(X) = 0 \times \frac{1}{12} + 1 \times \frac{1}{2} + 2 \times \frac{5}{12} = \frac{16}{12} = \frac{4}{3}\).