(i) There are 11 questions in total (8 from Part A and 3 from Part B). The number of ways to choose 6 questions from 11 is given by the combination formula:

\(\binom{11}{6} = 462\)

(ii) Candidates must choose at least 4 questions from Part A. We consider the following cases:

• 4 questions from Part A and 2 from Part B: \(\binom{8}{4} \times \binom{3}{2} = 70 \times 3 = 210\)
• 5 questions from Part A and 1 from Part B: \(\binom{8}{5} \times \binom{3}{1} = 56 \times 3 = 168\)
• 6 questions from Part A: \(\binom{8}{6} = 28\)

Adding these gives: \(210 + 168 + 28 = 406\)

(iii) Candidates must either choose both question 1 and question 2 in Part A, or choose neither. Consider the cases:

• Choose both question 1 and 2, then choose 4 more from the remaining 9 questions: \(\binom{9}{4} = 126\)
• Choose neither question 1 nor 2, then choose 6 from the remaining 9 questions: \(\binom{9}{6} = 84\)

Adding these gives: \(126 + 84 = 210\)