(a) To find the probability distribution of \(X\), we consider all possible sums of the numbers from the two spinners:

• Spinner 1: 0, 1, 2, 2
• Spinner 2: -1, 0, 1

Possible sums \(X\) are: -1, 0, 1, 2, 3.

Calculate probabilities:

• \(P(X = -1) = \frac{1}{12}\) (only one combination: 0 + (-1))
• \(P(X = 0) = \frac{2}{12}\) (combinations: 0 + 0, 1 + (-1))
• \(P(X = 1) = \frac{4}{12}\) (combinations: 0 + 1, 1 + 0, 2 + (-1), 2 + (-1))
• \(P(X = 2) = \frac{3}{12}\) (combinations: 1 + 1, 2 + 0, 2 + 0)
• \(P(X = 3) = \frac{2}{12}\) (combinations: 2 + 1, 2 + 1)

(b) To find \(\text{Var}(X)\), first calculate \(E(X)\):

\(E(X) = \sum x p(x) = -1 \times \frac{1}{12} + 0 \times \frac{2}{12} + 1 \times \frac{4}{12} + 2 \times \frac{3}{12} + 3 \times \frac{2}{12} = \frac{15}{12}\)

Then, calculate \(E(X^2)\):

\(E(X^2) = (-1)^2 \times \frac{1}{12} + 0^2 \times \frac{2}{12} + 1^2 \times \frac{4}{12} + 2^2 \times \frac{3}{12} + 3^2 \times \frac{2}{12} = \frac{35}{12}\)

Finally, \(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{35}{12} - \left( \frac{15}{12} \right)^2 = \frac{65}{48} = 1.35\).