(i) The tree diagram is as follows:

First draw two branches for the first sweet: Toffee (T) with probability \(\frac{6}{7}\) and Chocolate (C) with probability \(\frac{1}{7}\).

For the second sweet, if the first was a Toffee (T), draw branches for Toffee (T) with probability \(\frac{5}{9}\) and Chocolate (C) with probability \(\frac{4}{9}\).

If the first was a Chocolate (C), draw branches for Toffee (T) with probability \(\frac{6}{9}\) and Chocolate (C) with probability \(\frac{3}{9}\).

(ii) The probability distribution table for the number of toffees taken is:

(iii) The mean number of toffees taken is calculated as:

\(E(X) = \frac{90}{63} \times \left( \frac{10}{7} \right) = 1.43\)

(iv) The probability that the first sweet taken is a chocolate, given that the second sweet taken is a toffee, is:

\(P(\text{1st C | 2nd T}) = \frac{P(C \cap T)}{P(T)} = \frac{\frac{1}{7} \times \frac{6}{9}}{\frac{1}{7} \times \frac{6}{9} + \frac{6}{7} \times \frac{5}{9}} = \frac{6}{63} \div \frac{36}{63} = \frac{1}{6}\)