(i) The probability that a boy chooses Music is 0.15 (since 75% choose Games and 10% choose Drama, leaving 15% for Music). We use the binomial distribution with parameters: number of trials = 6, probability of success (choosing Music) = 0.15.

The probability that fewer than 3 boys choose Music is:

\(P(0, 1, 2) = (0.85)^6 + (0.15)(0.85)^5 \binom{6}{1} + (0.15)^2(0.85)^4 \binom{6}{2}\)

Calculating each term:

\((0.85)^6 = 0.377149515625\)

\((0.15)(0.85)^5 \binom{6}{1} = 0.2679075\)

\((0.15)^2(0.85)^4 \binom{6}{2} = 0.308\)

Summing these probabilities gives \(0.953\).

(ii) The probability that a student is a Drama student is:

\(P(D) = 0.6 \times 0.1 + 0.4 \times 0.55 = 0.28\)

The probability that a Drama student is a boy is:

\(P(B|D) = \frac{P(B \cap D)}{P(D)} = \frac{0.06}{0.28} = 0.2143\)

The probability that at least 1 of the 5 Drama students is a boy is:

\(P(>1) = 1 - P(0) = 1 - (0.7857)^5\)

\(P(0) = (0.7857)^5 = 0.7078\)

Thus, \(P(>1) = 1 - 0.7078 = 0.2922\)

Therefore, the probability is \(0.701\).