(a) The sum of probabilities must equal 1: \(p + r + 0.4 + 0.15 = 1\). Simplifying gives \(p + r + 0.55 = 1\), so \(p + r = 0.45\).

Given \(E(X) = 1.1\), we have \(0 \times 0.4 + 1 \times p + 2 \times r + 3 \times 0.15 = 1.1\). Simplifying gives \(p + 2r + 0.45 = 1.1\), so \(p + 2r = 0.65\).

Solving the equations \(p + r = 0.45\) and \(p + 2r = 0.65\) simultaneously, we subtract the first from the second: \((p + 2r) - (p + r) = 0.65 - 0.45\), giving \(r = 0.2\).

Substituting \(r = 0.2\) into \(p + r = 0.45\), we get \(p + 0.2 = 0.45\), so \(p = 0.25\).

(b) The variance \(\text{Var}(X)\) is calculated as \(E(X^2) - [E(X)]^2\).

First, calculate \(E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.25 + 2^2 \times 0.2 + 3^2 \times 0.15\).

\(E(X^2) = 0 + 0.25 + 0.8 + 1.35 = 2.4\).

Then, \(\text{Var}(X) = 2.4 - (1.1)^2 = 2.4 - 1.21 = 1.19\).