Okay so you're given the fact that f(x)=x^3+ax+6, when a is constant, has a factor of (x+3). This means that when it is divided by (x+3) the remainder is 0.
Part i). is to find the value of a, so if you do f(-3)=x^3+ax+6, (so substitute (x+3) into the polynomial) you will get:
f(-3) = (-3)^3 + (ax-3) + 6
f(-3) = -27 - 3a + 6
As (x+3) is a factor, f(-3) = 0, therefore:
0 = -27 - 3a + 6, meaning that:
27 = -3a + 6
21 = -3a
7 = -a
a = -7

Part ii). As we know that (x+3) is a factor from part ii, we divide the polynomial by (x+3), and substitute 'a', for -7 which we just found out in part i.
Divide it through long division, and you get:
x^3 - 7x + 6 divided by x+3 = x^2 - 3x + 2
So, as (x+3) is a factor, we know that if we were to multiply (x+3) by (x^2 - 3x + 2) we will get the original polynomial, and so we can rewrite it as:
(x+3)(x^2 - 3x + 2), x^2-3x+2 is a quadratic, and so we can factorise it simply, giving:
(x+3)(x-2)(x-1), giving values of x as -3, 2, and 1.
This has solved f(x)=0 as f(x)=0 means find the roots.
Hope this helped.