To find the probability distribution, we first determine \(k\) by ensuring the probabilities sum to 1:

\(k(-2)^2 + k(1)^2 + k(2)^2 + k(3)^2 = 1\)

\(4k + k + 4k + 9k = 18k = 1\)

\(k = \frac{1}{18}\)

Thus, the probability distribution table is:

To find \(E(X)\):

\(E(X) = \frac{4 \times (-2) + 1 \times 1 + 4 \times 2 + 9 \times 3}{18}\)

\(= \frac{-8 + 1 + 8 + 27}{18}\)

\(= \frac{28}{18} = \frac{14}{9}\)

To find \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{4 \times (-2)^2 + 1 \times 1^2 + 4 \times 2^2 + 9 \times 3^2}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{16 + 1 + 16 + 81}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{114}{18} - \frac{196}{81}\)

\(= \frac{317}{81}\)