(a) To find the probability of obtaining exactly 2 heads and 1 tail, consider the scenarios: HHT, HTH, and THH.

For HHT: \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{5} = \frac{4}{45}\)

For HTH: \(\frac{2}{3} \times \frac{1}{3} \times \frac{4}{5} = \frac{8}{45}\)

For THH: \(\frac{1}{3} \times \frac{2}{3} \times \frac{4}{5} = \frac{8}{45}\)

Total probability = \(\frac{4}{45} + \frac{8}{45} + \frac{8}{45} = \frac{20}{45} = \frac{4}{9}\)

(b) The probability distribution table for \(X\) is:

(c) Given \(\text{E}(X) = \frac{32}{15}\), find \(\text{Var}(X)\):

\(\text{Var}(X) = 0^2 \times \frac{1}{45} + 1^2 \times \frac{8}{45} + 2^2 \times \frac{20}{45} + 3^2 \times \frac{16}{45} - \left( \frac{32}{15} \right)^2\)

\(= \frac{8}{45} + \frac{80}{45} + \frac{144}{45} - \left( \frac{32}{15} \right)^2\)

\(= \frac{136}{225}\) or 0.604