(i) To satisfy the condition of having at least twice as many men as women, consider the following cases:

1. 4 men and 2 women: The number of ways is \(\binom{8}{4} \times \binom{4}{2} = 70 \times 6 = 420\).
2. 5 men and 1 woman: The number of ways is \(\binom{8}{5} \times \binom{4}{1} = 56 \times 4 = 224\).
3. 6 men and 0 women: The number of ways is \(\binom{8}{6} \times \binom{4}{0} = 28 \times 1 = 28\).

\(Total number of ways = 420 + 224 + 28 = 672.\)

(ii) First, calculate the total number of ways to choose 6 people from 12 (8 men + 4 women):

\(\binom{12}{6} = 924\).

Now, calculate the number of ways where the two particular men are together. Treat them as a single unit, so choose 4 more people from the remaining 10:

\(\binom{10}{4} = 210\).

Subtract the cases where the two men are together from the total:

\(924 - 210 = 714\).