(i) The probability distribution table is:

Since the total probability must equal 1, we have:

\(k + 2k + 3k + 4k = 1\)

\(10k = 1\)

Thus, \(k = \frac{1}{10}\).

(ii) The mean \(E(X)\) is calculated as:

\(E(X) = 1 \times \frac{1}{10} + 2 \times \frac{2}{10} + 3 \times \frac{3}{10} + 4 \times \frac{4}{10}\)

\(E(X) = \frac{1}{10} + \frac{4}{10} + \frac{9}{10} + \frac{16}{10} = 3\)

The variance \(\text{Var}(X)\) is calculated as:

\(\text{Var}(X) = \left(1^2 \times \frac{1}{10} + 2^2 \times \frac{2}{10} + 3^2 \times \frac{3}{10} + 4^2 \times \frac{4}{10}\right) - E(X)^2\)

\(\text{Var}(X) = \left(\frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10}\right) - 3^2\)

\(\text{Var}(X) = \frac{100}{10} - 9 = 1\)