(a) To find the probabilities, we use \(P(X = x) = k(x^2 - 1)\) for each value of \(x\):

• For \(x = -2\), \(P(X = -2) = k((-2)^2 - 1) = 3k\).
• For \(x = 2\), \(P(X = 2) = k(2^2 - 1) = 3k\).
• For \(x = 3\), \(P(X = 3) = k(3^2 - 1) = 8k\).

Since the total probability must be 1, we have:

\(3k + 3k + 8k = 1\)

\(14k = 1\)

\(k = \frac{1}{14}\)

Thus, the probability distribution table is:

(b) To find \(E(X)\), we calculate:

\(E(X) = (-2) \times \frac{3}{14} + 2 \times \frac{3}{14} + 3 \times \frac{8}{14}\)

\(= \frac{-6}{14} + \frac{6}{14} + \frac{24}{14}\)

\(= \frac{12}{14} = \frac{12}{7}\)

To find \(\text{Var}(X)\), we use:

\(\text{Var}(X) = (-2)^2 \times \frac{3}{14} + 2^2 \times \frac{3}{14} + 3^2 \times \frac{8}{14} - (E(X))^2\)

\(= 4 \times \frac{3}{14} + 4 \times \frac{3}{14} + 9 \times \frac{8}{14} - \left(\frac{12}{7}\right)^2\)

\(= \frac{12 + 12 + 72}{14} - \frac{144}{49}\)

\(= \frac{96}{14} - \frac{144}{49}\)

\(= \frac{192}{49}\)