(a) To find the probability that Sadie takes exactly 1 red ball, we calculate the probability of choosing 1 red ball and 2 blue balls. The probability is given by:

\(P(1 \text{ red}) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} = \frac{15}{56}\)

(b) The probability distribution table for X is:

(c) To find \(\text{Var}(X)\), we use the formula:

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

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

\(E(X^2) = 0^2 \times \frac{1}{56} + 1^2 \times \frac{15}{56} + 2^2 \times \frac{30}{56} + 3^2 \times \frac{10}{56}\)

\(E(X^2) = \frac{15}{56} + \frac{120}{56} + \frac{90}{56} = \frac{225}{56}\)

Now, substitute into the variance formula:

\(\text{Var}(X) = \frac{225}{56} - \left( \frac{15}{8} \right)^2\)

\(\text{Var}(X) = \frac{225}{56} - \frac{225}{64}\)

\(\text{Var}(X) = 0.502\)