(a) The probability distribution table for X is:

To find k, use the fact that the sum of probabilities is 1:

\(4k + 6k + 6k + 4k = 1\)

\(20k = 1\)

\(k = \frac{1}{20}\)

(b) Calculate the expected value \(E(X)\):

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

\(= \frac{4}{20} + \frac{12}{20} + \frac{18}{20} + \frac{16}{20}\)

\(= \frac{50}{20} = 2.5\)

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

\(\text{Var}(X) = 1^2 \times \frac{4}{20} + 2^2 \times \frac{6}{20} + 3^2 \times \frac{6}{20} + 4^2 \times \frac{4}{20} - (2.5)^2\)

\(= \left(\frac{4}{20} + \frac{24}{20} + \frac{54}{20} + \frac{64}{20}\right) - 6.25\)

\(= \frac{146}{20} - 6.25\)

\(= 7.3 - 6.25\)

\(= 1.05\)