(a) To choose a group of 6 with exactly 1 man, we select 1 man from 4 men and 5 women from 11 women. The number of ways to do this is given by:

\(\binom{4}{1} \times \binom{11}{5} = 4 \times 462 = 1848\)

(b) We need to find the number of ways to choose a group of 6 such that Jane and Kate are not both included. We can use two methods:

Method 1: Consider the scenarios:

1. Neither Jane nor Kate is selected: Choose 6 from the remaining 13 people.
2. Only Jane is selected: Choose 5 from the remaining 13 people.
3. Only Kate is selected: Choose 5 from the remaining 13 people.

The total number of ways is:

\(\binom{13}{6} + \binom{13}{5} + \binom{13}{5} = 1716 + 1287 + 1287 = 4290\)

Method 2: Calculate the total number of ways to choose 6 people from 15, then subtract the number of ways where both Jane and Kate are included:

\(\binom{15}{6} - \binom{13}{4} = 5005 - 715 = 4290\)