Question 526662
When you divide a polynomial P(x) by a divisor polynomial D(x) you get a quotient polynomial Q(x) and a remainder polynomial R(x). With some luck, all of those polynomial will be very small and simple, and maybe the remainder will be 0, or just a number (a polynomial of degree zero).

1) {{{P(x)=x^3+mx^2-nx-6=(x-2)Q(x)}}} (remainder is zero) means that
{{{P(2)=2^3+m(2^2)-2n-6=(2-2)Q(2)=0}}}
{{{8+4m-2n-6=0}}} or {{{4m-2n=-2}}}
Similarly
{{{P(-3)=0}}} leads you to {{{9m+3n=33}}}
Those two equations form a linear system that allows you to find m and n.

2) Same idea
{{{P(x)=(x-3)Q(x)}}} (remainder is zero) so {{{P(3)=0}}} and
{{{P(x)=(x+1)Q(x)+8}}}  (remainder is 8) so {{{P(-1)=8}}}
You'll get two equations that would allow you to find a and b.

3) {{{P(x)=(x^2-4)Q(x)=(x-2)(x+2)Q(x)}}} (remainder is zero)
I would bet on
{{{Q(x)=2x-1}}} and {{{P(x)=(x^2-4)(2x-1)}}}
Multiplying will tell you if that product is a polynomial that could be the one you were given. (Easy way to solve it).
You are probably expected to use
{{{P(2)=0}}} and {{{P(-2)=0}}}
to set up a system of equations to solve for a and b, but who really enjoys solving systems of equations?