(i) Total number of selections = \(\binom{12}{7} = 792\).

Selections with the boy included = \(\binom{11}{6} = 462\).

Probability = \(\frac{462}{792} = \frac{7}{12}\).

(ii) Method 1: Scenarios are:

• 2G + 5B: \(\binom{4}{2} \times \binom{8}{5} = 336\)
• 3G + 4B: \(\binom{4}{3} \times \binom{8}{4} = 280\)
• 4G + 3B: \(\binom{4}{4} \times \binom{8}{3} = 56\)

\(Total = 672.\)

Probability = \(\frac{672}{792} = \frac{28}{33}\).

Let \(C\) represent completing the puzzle and \(C'\) represent not completing it.

The probability that Kenny completes the puzzle on both days is \(P(C, C) = 0.8 \times 0.9 = 0.72\).

The probability that Kenny completes the puzzle on Monday but not on Tuesday is \(P(C, C') = 0.8 \times 0.1 = 0.08\).

The probability that Kenny does not complete the puzzle on Monday but completes it on Tuesday is \(P(C', C) = 0.2 \times 0.6 = 0.12\).

The probability that Kenny completes the puzzle on at least one of the two days is:

\(P(C, C) + P(C, C') + P(C', C) = 0.72 + 0.08 + 0.12 = 0.92\)

Alternatively, the probability that Kenny does not complete the puzzle on both days is \(P(C', C') = 0.2 \times 0.4 = 0.08\).

Thus, the probability that he completes the puzzle on at least one day is:

\(1 - P(C', C') = 1 - 0.08 = 0.92\)