(a) To have more women than men, consider the following scenarios:

1. 6 women and 0 men: \(\binom{9}{6} = 84\)
2. 5 women and 1 man: \(\binom{9}{5} \times \binom{5}{1} = 126 \times 5 = 630\)
3. 4 women and 2 men: \(\binom{9}{4} \times \binom{5}{2} = 126 \times 10 = 1260\)

\(Total number of ways = 84 + 630 + 1260 = 1974.\)

(b) Total number of ways to choose 6 people from 14: \(\binom{14}{6} = 3003\).

Number of ways with both sister and brother: \(\binom{12}{4} = 495\).

Number of ways without both sister and brother: \(3003 - 495 = 2508\).

Alternatively, calculate:

1. Number of ways with neither: \(\binom{12}{6} = 924\)
2. Number of ways with either brother or sister (not both): \(\binom{12}{5} \times 2 = 792 \times 2 = 1584\)

\(Total number of ways = 924 + 1584 = 2508.\)