(i) The probability of taking a toffee from Susan's bag is \(\frac{5}{12}\). After placing it in Ahmad's bag, the probability of taking a boiled sweet from Ahmad's bag is \(\frac{2}{10}\). Therefore, the probability that the two sweets taken are a toffee from Susan’s bag and a boiled sweet from Ahmad’s bag is:

\(P(T, B) = \frac{5}{12} \times \frac{2}{10} = \frac{1}{12}\)

(ii) Let \(C_S\) be the event that a chocolate is taken from Susan's bag, and \(C_A\) be the event that a chocolate is taken from Ahmad's bag. We need to find \(P(C_S | C_A)\).

First, calculate \(P(C_S \cap C_A)\):

\(P(C_S \cap C_A) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120}\)

Next, calculate \(P(C_A)\):

\(P(C_A) = \frac{7}{12} \times \frac{4}{10} + \frac{5}{12} \times \frac{3}{10} = \frac{28}{120} + \frac{15}{120} = \frac{43}{120}\)

Finally, calculate \(P(C_S | C_A)\):

\(P(C_S | C_A) = \frac{P(C_S \cap C_A)}{P(C_A)} = \frac{28/120}{43/120} = \frac{28}{43}\)