Answer: \(27720\), \(120\), \(48\), \(42\).

(a) Choose the group of \(3\), then the group of \(4\), and the remaining \(5\) people form the last group:

\(\binom{12}{3}\binom{9}{4}=220\times126=27720.\)

(b)(i) The number of four-digit numbers is

\({}^5P_4=5\times4\times3\times2=120.\)

(b)(ii) For the number to be even, the last digit must be \(2\) or \(8\). Choose the last digit first, then arrange the remaining three positions:

\(2\times4\times3\times2=48.\)

(b)(iii) The number must be greater than \(7000\) and odd.

If the first digit is \(7\) or \(9\), there are \(2\) choices for the first digit, \(2\) choices for the odd final digit, and then \(3\times2\) ways to fill the middle positions:

\(2\times2\times3\times2=24.\)

If the first digit is \(8\), the final digit can be \(3\), \(7\) or \(9\), and the two middle positions can be filled in \(3\times2\) ways:

\(3\times3\times2=18.\)

So the total is

\(24+18=42.\)