(i) To have more girls than boys in the team of 5, the possible combinations are:

• 5 girls and 0 boys: \(\binom{8}{5} \times \binom{6}{0} = 56\)
• 4 girls and 1 boy: \(\binom{8}{4} \times \binom{6}{1} = 420\)
• 3 girls and 2 boys: \(\binom{8}{3} \times \binom{6}{2} = 840\)

\(Total = 56 + 420 + 840 = 1316\)

(ii) If three of the boys are cousins and are either all in the team or all not in the team, consider two cases:

• All cousins in the team: Choose 2 more from the remaining 11 (3 boys + 8 girls): \(\binom{11}{2} = 55\)
• All cousins not in the team: Choose 5 from the remaining 11 (3 boys + 8 girls): \(\binom{11}{5} = 462\)

\(Total = 55 + 462 = 517\)