I did the first part successfully:
$$w^{12}=1= \cos 2\pi + i \sin 2\pi$$
$$w= \cos \frac{\pi k}{6} + i \sin \frac{\pi k}{6}$$
Where $k=0,1,2,3,4,5,6,7,8,9,10,11$
I struggled with this for a long time, I know that the two real roots are $z=2$ and $z=-2$
But how do I prove it and show it can be expressed as that. Please help.
Im new to these types of questions
I only know how to do this when its something like:
$$w^4=1$$
Where there is a the number $1$ or $-1$ and one sided equation. Because then I can write
$$w^4-1=0$$
$$(w-1)(w+1)(w^2+1)=0$$
By writing the second equation as $$\Bigl(\frac{z+2}{z}\Bigr)^{12}=1$$ it reduces to the first. Therefore the solutions are $$\frac{z+2}{z}=\cos\frac{k\pi}{6}+i\sin\frac{k\pi}{6}\tag{$*$}$$ for $k=0$ or $k=6$ or $k=\pm1,\pm2,\ldots,\pm5$. Now $k=0$ gives $$\frac{z+2}{z}=1$$ which is impossible, and $k=6$ gives $$\frac{z+2}{z}=-1$$ which has the real solution $z=-1$. So this leaves ten non-real solutions. You can find them from $(*)$ and do a bit of simplification using trig functions in order to arrive at the given answers. Specifically, solve for $z$, rationalise the denominator and use half-angle formulae: $$\eqalign{z &=-\frac{2}{1-(\cos\frac{k\pi}{6}+i\sin\frac{k\pi}{6})}\cr &=-2\frac{(1-\cos\frac{k\pi}{6})+i\sin\frac{k\pi}{6}} {(1-\cos\frac{k\pi}{6})^2+(\sin\frac{k\pi}{6})^2}\cr &=\cdots\cr &=-\frac{(1-\cos\frac{k\pi}{6})+i\sin\frac{k\pi}{6}}{1-\cos\frac{k\pi}{6}} \cr &=-1-i\frac{2\sin\frac{k\pi}{12}\cos\frac{k\pi}{12}}{2\sin^2\frac{k\pi}{12}} \ .\cr}$$ In fact, if you are familiar with complex exponentials you can write $(*)$ as $$\frac{z+2}{z}=e^{k\pi i/6}$$ and the algebra becomes much easier.
Comment. Worth a thought: including the real solution, we have $11$ solutions. Can you explain why this is so, when your original equation involved polynomials of degree $12$? Why are there not $12$ solutions?