(a) To find \(k\), use the fact that the sum of probabilities must equal 1:

\(2k + 6k + 12k + 20k = 1\)

\(40k = 1\)

\(k = \frac{1}{40}\)

Now, calculate the probabilities:

(b) Calculate \(E(X)\):

\(E(X) = \frac{1 \times 2 + 2 \times 6 + 3 \times 12 + 4 \times 20}{40}\)

\(E(X) = \frac{2 + 12 + 36 + 80}{40}\)

\(E(X) = \frac{130}{40} = 3.25\)

Calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{1^2 \times 2 + 2^2 \times 6 + 3^2 \times 12 + 4^2 \times 20}{40} - (E(X))^2\)

\(\text{Var}(X) = \frac{2 + 24 + 108 + 320}{40} - (3.25)^2\)

\(\text{Var}(X) = \frac{454}{40} - 10.5625\)

\(\text{Var}(X) = 11.35 - 10.5625 = 0.7875\)