(a) Let the probability of a person being a Notes supporter be 0.3. We use the binomial distribution with parameters 6 (trials) and 0.3 (success probability).

The probability that no more than 2 people are Notes supporters is:

\(P(0, 1, 2) = \binom{6}{0} (0.3)^0 (0.7)^6 + \binom{6}{1} (0.3)^1 (0.7)^5 + \binom{6}{2} (0.3)^2 (0.7)^4\)

Calculating each term:

\(\binom{6}{0} (0.3)^0 (0.7)^6 = 0.1176\)

\(\binom{6}{1} (0.3)^1 (0.7)^5 = 0.3025\)

\(\binom{6}{2} (0.3)^2 (0.7)^4 = 0.3241\)

Adding these probabilities gives:

\(0.1176 + 0.3025 + 0.3241 = 0.744\)

(b) The probability that a person supports neither choir is:

\(1 - (0.3 + 0.45) = 0.25\)

The probability that none of the 6 people support either choir is:

\(P(6 \text{ support neither choir}) = (0.25)^6\)

Calculating this gives:

\((0.25)^6 = 0.000244\)