(i) To find the probability of obtaining exactly 1 head and 2 tails, consider the scenarios: HTT, THT, and TTH.

For HTT: \(P(H) = \frac{2}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(HTT) = \frac{2}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For THT: \(P(T) = \frac{1}{3}, P(H) = \frac{2}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(THT) = \frac{1}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For TTH: \(P(T) = \frac{1}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(H) = \frac{1}{4}\) for Coin B.

\(P(TTH) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{36}\).

Total probability: \(\frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{12}{36} + \frac{1}{36} = \frac{13}{36}\).

(ii) The probability distribution table is:

(iii) The expectation \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{3}{36} + 1 \times \frac{13}{36} + 2 \times \frac{16}{36} + 3 \times \frac{4}{36}\).

\(E(X) = \frac{0}{36} + \frac{13}{36} + \frac{32}{36} + \frac{12}{36} = \frac{57}{36} = \frac{19}{12} \approx 1.58\).