(a) To form a committee of 6 people with 3 men and 3 women from 6 men and 8 women, we use combinations:

The number of ways to choose 3 men from 6 is given by \(^6C_3\).

The number of ways to choose 3 women from 8 is given by \(^8C_3\).

Thus, the total number of ways is \(^6C_3 \times ^8C_3 = 20 \times 56 = 1120\).

(b) We need to form a committee of 6 people with no more than 2 brothers. We consider the following cases:

1. 0 brothers: Choose 0 brothers from 3 and 6 others from the remaining 11 people (3 non-brothers + 8 women): \(^3C_0 \times ^{11}C_6 = 1 \times 462 = 462\).
2. 1 brother: Choose 1 brother from 3 and 5 others from the remaining 11 people: \(^3C_1 \times ^{11}C_5 = 3 \times 462 = 1386\).
3. 2 brothers: Choose 2 brothers from 3 and 4 others from the remaining 11 people: \(^3C_2 \times ^{11}C_4 = 3 \times 330 = 990\).

Adding these cases gives the total number of ways: \(462 + 1386 + 990 = 2838\).