(a) To find \(P(B)\), consider the scenarios where at least one biased coin shows a head. The probability that a biased coin shows a tail is \(\frac{3}{4}\). The probability that both biased coins show tails is \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\). Therefore, \(P(B) = 1 - \frac{9}{16} = \frac{7}{16}\).

(b) To find \(P(A \mid B)\), use the formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\). The event \(A \cap B\) occurs when all coins show heads or all show tails. The probability of all heads is \(\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}\). The probability of all tails is \(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{32}\). Thus, \(P(A \cap B) = \frac{1}{32}\). Therefore, \(P(A \mid B) = \frac{\frac{1}{32}}{\frac{7}{16}} = \frac{1}{14}\).

(c) The probability distribution for \(X\) is calculated as follows:

• \(P(X = 0) = \frac{9}{32}\)
• \(P(X = 1) = \frac{15}{32}\)
• \(P(X = 2) = \frac{7}{32}\)
• \(P(X = 3) = \frac{1}{32}\)