(a) For Box A, the probability of choosing a red ball is \(\frac{7}{8}\) and a blue ball is \(\frac{1}{8}\).

For Box B, if a red ball is added, the probability of choosing a red ball is \(\frac{10}{15}\) and a blue ball is \(\frac{5}{15}\). If a blue ball is added, the probability of choosing a red ball is \(\frac{9}{15}\) and a blue ball is \(\frac{6}{15}\).

(b) The probability that the two balls chosen are not the same colour is calculated as:

\(\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{9}{15} = \frac{44}{120} = \frac{11}{30}\)

(c) The probability that the ball chosen from box A is blue given that the ball chosen from box B is blue is:

\(P(A \text{ blue } | B \text{ blue }) = \frac{P(A \text{ blue } \cap B \text{ blue })}{P(B \text{ blue })}\)

\(= \frac{\frac{1}{8} \times \frac{6}{15}}{\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{6}{15}} = \frac{\frac{1}{20}}{\frac{41}{120}} = \frac{6}{41}\)