(iii) To have more males than females in a group of 5, we can have the following combinations:

1. 3 males and 2 females: The number of ways to choose 3 males from 5 is \(\binom{5}{3}\), and the number of ways to choose 2 females from 10 is \(\binom{10}{2}\). Thus, the total number of ways is \(\binom{5}{3} \times \binom{10}{2} = 450\).
2. 4 males and 1 female: The number of ways to choose 4 males from 5 is \(\binom{5}{4}\), and the number of ways to choose 1 female from 10 is \(\binom{10}{1}\). Thus, the total number of ways is \(\binom{5}{4} \times \binom{10}{1} = 50\).
3. 5 males and 0 females: The number of ways to choose 5 males from 5 is \(\binom{5}{5}\), and the number of ways to choose 0 females from 10 is \(\binom{10}{0}\). Thus, the total number of ways is \(\binom{5}{5} \times \binom{10}{0} = 1\).

Adding these, the total number of ways is \(450 + 50 + 1 = 501\).

(iv) Mr and Mrs Blake are already chosen, so we need to choose 3 more performers. We need more males than females, so the combinations are:

1. 3 males and 0 females: Choose 3 males from the remaining 4 males. The number of ways is \(\binom{4}{3} \times \binom{9}{0} = 4\).
2. 2 males and 1 female: Choose 2 males from the remaining 4 males and 1 female from 9 females. The number of ways is \(\binom{4}{2} \times \binom{9}{1} = 54\).

Adding these, the total number of ways is \(4 + 54 = 58\).