Determine the cube roots of -8 in polar form Exam time tomorrow and I am not entirely sure if I am doing this right.
I first write -8 as a complex number
$z^3 = -8 = -8 \times 0i$
Calculate the modulus of z
$|z| = \sqrt{-8^2} = 8$
Get the arg of z
$tan^{-1} = \frac{0}{-8} = 0 = \pi$
Write the number in polar form
$\theta = \pi + 2k\pi$
$z^3 = 8(cos(\pi + 2k\pi) + isin(\pi + 2k\pi))$  
Use De Moivre's theorem
$z = 8^\frac{1}{3}(cos(\pi + 2k\pi) + isin(\pi + 2k\pi))^\frac{1}{3}\\
= 2(cos(\frac{\pi}{3} + \frac{2k\pi}{3}) + isin(\frac{\pi}{3} + \frac{2k\pi}{3}))$
and then I fill out the equation with different values of k
$
k = 0, z= 2(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))\\
k = 1, z= 2(cos(\pi) + isin(\pi))\\
k = 2, z= 2(cos(\frac{5\pi}{3}) + isin(\frac{5\pi}{3}))
$
Is this correct?

Looks good so far. 
What you'll also likely want to do is evaluate $\cos \theta, \sin\theta$ for each $\theta.$ The trigonometric values needed here are for angles that are fairly basic, so you'll want to re-familiarize yourself with the standard angles, expressed in radians, and the corresponding trig values with those angles as arguments. See this link, and the image below, both from Wikipedia.
Taking your final work, but evaluating the $\sin$ and $\cos$ of the angles gives us an explicit representation of the cube roots of $-8$:
$$
k = 0, z= 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 2\left(\frac 12 + i\frac {\sqrt 3}{2}\right) = 1 + i\sqrt 3\\
k = 1, z= 2(\cos(\pi) + i\sin(\pi)) =  2\left(-1 + i\cdot 0\right) = -2\\
k = 2, z= 2(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}))= 2\left(\frac 12 - i\frac{\sqrt 3}{2}\right) = 1 - i\sqrt 3
$$
Trigonometric values of standard angles, using the unit-circle definition:

