I'm given that complex number and I'm asked to calculate $\omega^5$, which I guess it's $\left(e^{2i\pi/5}\right)^5=e^{2i\pi}=1$. Then, I have to prove $1+\omega+\omega^2+\omega^3+\omega^4=0$. I've done it like that:
$$\omega·(1+\omega+\omega^2+\omega^3+\omega^4)=\omega+\omega^2+\omega^3+\omega^4+\omega^5=1+\omega+\omega^2+\omega^3+\omega^4$$
Therefore, since $\omega\neq 1$, we have $1+\omega+\omega^2+\omega^3+\omega^4=0$.
Now I'm asked to find a quadratic equation with integer coefficients with roots $\omega+\omega^4$ and $\omega^2+\omega^3$. Here is my try:
$$(x-(\omega+\omega^4))·(x-(\omega^2+\omega^3))=(x-\omega-\omega^4)(x-\omega^2-\omega^3)$$
Since $-\omega^2-\omega^3=1+\omega+\omega^4$,
$$(x-\omega-\omega^4)(x+\omega^4+\omega+1)=x^2+x-2-\omega-\omega^2-\omega^3-\omega^4=0$$
Therefore, the quadratic equation would be $x^2+x-1=0$. Is there any mistake? I'm not sure about my answer.
You seem to have written a perfectly valid proof. Just note that the general case is also true (and pretty important as well). Let $n\in\mathbb{N}$, and consider the complex $n$th root of $1$, i.e., the complex number $\omega_n=e^{\frac{2\pi i}{n}}=\cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$ that satisfies the equation $z^n=1$. Of course there are $n$ solutions to this equation, we take $\omega_n$ as the one with the minimal argument (except from $1$). The calculation of $\omega_n^n$ is as you have done - using de Moivre formula, $\omega_n^n=(e^{\frac{2\pi i}{n}})^n=e^{2\pi i}=1$.
By the sum of a geometric sequence (which is also true for complex numbers), we know that $$1+z+z^2+\ldots+z^{n-1}=\frac{z^n-1}{z-1},$$ and therefore plugging in $z=\omega_n$ yields zero in the RHS, and the sum you were requested to calculate in the LHS.