Answer: (a)(i) \(15120\); (a)(ii) \(3780\); (b) \(116\).

We start with the main method. Use permutations when order matters, and separate cases when the first or last digit has restrictions.

(a)(i) A 5-digit number is formed from 9 digits without repetition, so the number of possibilities is

\({}^{9}P_5=9\times8\times7\times6\times5=15120.\)

(a)(ii) The number must be greater than \(50000\) and even.

If the first digit is \(5,7,\) or \(9\), the final digit can be any of \(2,4,6,8\). Then the three middle digits can be arranged in \({}^{7}P_3=210\) ways. This gives

\(3\times4\times210=2520.\)

If the first digit is \(6\) or \(8\), the final digit must be even and different from the first digit, so there are \(3\) choices for the final digit. The three middle digits can again be arranged in \(210\) ways. This gives

\(2\times3\times210=1260.\)

Hence the total is

\(2520+1260=3780.\)

(b) Since there are only 2 nurses and the team must include at least 2 nurses, both nurses must be chosen.

We then choose 7 people from doctors and dentists, with at least 2 doctors and at least 2 dentists.

The possible cases are:

\(5\) doctors and \(2\) dentists: \(\binom65\binom42=6\cdot6=36.\)

\(4\) doctors and \(3\) dentists: \(\binom64\binom43=15\cdot4=60.\)

\(3\) doctors and \(4\) dentists: \(\binom63\binom44=20\cdot1=20.\)

So the number of teams is

\(36+60+20=116.\)

This completes the solution and gives the required result.