How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$ 

I did the first parts : 
$$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$

$\omega=-1$ is a root so : 
$$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$
$$\omega^4-\omega^3+\omega^2-\omega=-1$$

$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (\cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5})$$
$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (-\cos \frac{\pi}{5} + i \sin \frac{\pi}{5})$$
$$\omega-\omega^4= 2 \cos \frac{\pi}{5}$$
Did similar method for $\omega^3-\omega^2$ and got the desired result.

For this part I only know the first part which is: 
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}$$
$$2\cos {\pi}{5} + 2\cos \frac{3\pi}{5}=\omega-\omega^4+\omega^3-\omega^2=1$$
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}=\frac{1}{2}$$

$$\cos \frac{\pi}{5} \cos \frac{3\pi}{5} = \frac{1}{4} (\omega-\omega^4)(\omega^3-\omega^2)$$
$$\frac{1}{4} (\omega^4-\omega^3-\omega^7+\omega^6)$$
How do I find $\omega^7$ and $\omega^6$ ???? please help.

Also I would love if I receive support for the last part too, I don't know how  to make a quadratic equation

Since $\omega^5+1=0$, you have that $\omega^5=-1$, so $\omega^6=\omega^5\cdot\omega=-\omega$ etc.
For the quadratic equation, use that $(x-a)(x-b)=x^2-(a+b)x+ab$, let $a=\frac{1}{2}(\omega-\omega^4)=\cos(\pi/5)$ and $b=\frac{1}{2}(\omega^3-\omega^2)$. You calculated values for $a+b$ and $ab$ just beforehand, so you have the quadratic equation (namely $x^2-x-1$).
