(a) The sum of probabilities must equal 1:

\(p + q + 0.6 + 0.05 = 1\)

\(p + q + 0.65 = 1\)

\(p + q = 0.35\) (1)

Using the expected value \(E(X) = 0.55\):

\(0 \times 0.6 + 1 \times p + 2 \times q + 3 \times 0.05 = 0.55\)

\(p + 2q + 0.15 = 0.55\)

\(p + 2q = 0.4\) (2)

Solving equations (1) and (2):

From (1): \(p = 0.35 - q\)

Substitute into (2):

\(0.35 - q + 2q = 0.4\)

\(0.35 + q = 0.4\)

\(q = 0.05\)

Substitute \(q = 0.05\) into (1):

\(p + 0.05 = 0.35\)

\(p = 0.3\)

(b) The variance \(\text{Var}(X)\) is given by:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

Calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times 0.6 + 1^2 \times 0.3 + 2^2 \times 0.05 + 3^2 \times 0.05\)

\(E(X^2) = 0 + 0.3 + 0.2 + 0.45\)

\(E(X^2) = 0.95\)

Thus, \(\text{Var}(X) = 0.95 - 0.55^2\)

\(\text{Var}(X) = 0.95 - 0.3025\)

\(\text{Var}(X) = 0.6475\)