Answer: (a) \(6435\); (b) \(9\); (c) \(1425\).

We start with the main method. Use combinations when order does not matter, and split the count into valid cases if there are lower bounds.

There are \(6+5+4=15\) people altogether.

For part (a), choose \(8\) people from \(15\):

\(\binom{15}{8}=6435.\)

For part (b), without teachers means choose all \(8\) people from the \(5\) doctors and \(4\) police officers.

There are \(9\) non-teachers, so

\(\binom98=9.\)

For part (c), the team has \(8\) people and the number of doctors must equal the number of teachers.

If there are \(2\) teachers and \(2\) doctors, then there must be \(4\) police officers:

\(\binom62\binom52\binom44=15\cdot10\cdot1=150.\)

If there are \(3\) teachers and \(3\) doctors, then there must be \(2\) police officers:

\(\binom63\binom53\binom42=20\cdot10\cdot6=1200.\)

If there are \(4\) teachers and \(4\) doctors, then there are no police officers:

\(\binom64\binom54\binom40=15\cdot5\cdot1=75.\)

Therefore the total number of teams is

\(150+1200+75=1425.\)

This completes the solution and gives the required result.