(i) The probability distribution table is:

Since the sum of probabilities must equal 1, we have:

\(k + k + 4k + 9k = 15k = 1\)

Solving for \(k\), we get \(k = \frac{1}{15}\).

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

\(E(X) = (-1)k + 1k + 2(4k) + 3(9k) = -k + k + 8k + 27k = 35k\)

Substituting \(k = \frac{1}{15}\), we get:

\(E(X) = 35 \times \frac{1}{15} = \frac{35}{15} = \frac{7}{3}\)

To find \(Var(X)\), we use:

\(Var(X) = E(X^2) - (E(X))^2\)

\(E(X^2) = (-1)^2k + 1^2k + 2^2(4k) + 3^2(9k) = k + k + 16k + 81k = 99k\)

\(Var(X) = 99k - \left(\frac{7}{3}\right)^2\)

Substituting \(k = \frac{1}{15}\), we get:

\(Var(X) = 99 \times \frac{1}{15} - \left(\frac{7}{3}\right)^2 = \frac{99}{15} - \frac{49}{9}\)

\(Var(X) = \frac{52}{45}\)