Answer: (a)(i) \(3003\), (a)(ii) \(858\), (b) \(68880\).

Count the choices in stages, applying any restrictions before multiplying the number of possibilities together.

Without restrictions, the number of teams is

\({15\choose10}=3003.\)

If the 3 sisters must not be separated, either all 3 sisters are selected or none of them are selected.

If all 3 sisters are selected, choose the remaining 7 players from the other 12:

\({12\choose7}=792.\)

If none of the sisters are selected, choose all 10 players from the other 12:

\({12\choose10}=66.\)

So the total is

\(792+66=858.\)

For the even 6-digit numbers, consider the final digit.

If the final digit is \(0\), then the first digit has 9 choices, and the remaining four positions have \({}^{8}P_4\) choices. This gives

\(9\times{}^{8}P_4=9\times8\times7\times6\times5=15120.\)

If the final digit is one of \(2,4,6,8\), there are 4 choices for the final digit. The first digit can be any non-zero digit except the final digit, so it has 8 choices. The remaining four positions have \({}^{8}P_4\) choices:

\(4\times8\times{}^{8}P_4=4\times8\times8\times7\times6\times5=53760.\)

Therefore the total is

\(15120+53760=68880.\)

This gives the required answer: (a)(i) \(3003\), (a)(ii) \(858\), (b) \(68880\).