(i) To have more women than men, the possible combinations are:

• 4 women and 2 men: \(\binom{8}{4} \times \binom{5}{2} = 700\)
• 5 women and 1 man: \(\binom{8}{5} \times \binom{5}{1} = 280\)
• 6 women and 0 men: \(\binom{8}{6} \times \binom{5}{0} = 28\)

\(Total ways = 700 + 280 + 28 = 1008\)

(ii) For 3 men and 3 women, but two particular men refuse to be together:

• Choose 3 men from 5, and 3 women from 8 without restriction: \(\binom{5}{3} \times \binom{8}{3} = 560\)
• Subtract the cases where the two particular men are together:
• Choose 2 men (including the two particular men) and 3 women: \(\binom{3}{2} \times \binom{8}{3} = 168\)

\(Total ways = 560 - 168 = 392\)