To find the values of \(p\) and \(q\), we use the following equations:

1. The sum of probabilities: \(0.08 + p + 0.12 + 0.16 + q + 0.22 = 1\)

2. The mean equation: \(-2(0.08) - 1(p) + 0(0.12) + 1(0.16) + 2(q) + 3(0.22) = 1.05\)

Simplifying these equations:

\(p + q = 0.42\)

\(-0.16 - p + 0.16 + 2q + 0.66 = 1.05\)

\(-p + 2q = 0.39\)

Solving these equations simultaneously:

\(p = 0.15\)

\(q = 0.27\)

To find the variance of \(X\), we use the formula:

\(\text{Var}(X) = \sum x^2 P(X = x) - (\text{mean})^2\)

\(\text{Var}(X) = 4(0.08) + p + 0.16 + 4q + 1.98 - (1.05)^2\)

Substituting \(p = 0.15\) and \(q = 0.27\):

\(\text{Var}(X) = 2.59\)