Part 1 is still kind of nice; depending on what you need this for this may or may not be useful:
For a real polynomial, its roots are either real or form conjugate pairs. If you have two conjugate roots of unity
$$e^{\frac{2i\pi k}{n}},\ e^{-\frac{2i\pi k}{n}},$$
then the quadratic polynomial with them as roots is
$$P(x)=x^2-2\cos\left(\frac{2\pi k}{n}\right)x+1.$$
So, any real-coefficiented polynomial (whose roots are all roots of unity) has the form
$$(x+1)^m(x-1)^n\prod_{k=1}^N \left(x^2-2\cos\left(\frac{2\pi a_k}{b_k}\right)x+1\right),$$
where $m,n$ are nonnegative integers and all $a_k,b_k$ are positive integers (with $a_k<b_k$, if you wish).
For part 2, I mean, you can just pick the roots to be
$$x_k=e^{\frac{2i\pi a_k}{b_k}},$$
and your polynomial will be
$$\prod_{k=1}^N \left(x-e^{\frac{2i\pi a_k}{b_k}}\right),$$
but this probably isn't that useful. Another thing that may or may not be useful for either of these is that all polynomials with only roots of unity as roots have roots that all satisfy $x^M=1$ for some (possibly large with respect to the degree of the polynomial) integer $M$. In the case above this $M$ can be taken to be $b_1b_2\cdots b_k$. So, all polynomials with only roots of unity as roots are factors of $x^M-1$ for some $M$.
If a polynomial $\phi$ has a repeated root $a$ then $\phi(X)=(X-a)^k \psi(X)$ for some polynomial $\psi$ and some $k\geqslant 2$. Then $\phi'(X)=k(X-a)^{k-1}\psi(X)+(X-a)^k \psi'(X)$, and $\phi'(a)=0$.
Apply this to $X^n-1$ whose derivative is $nX^{n-1}$: the only candidate for a repeated root is $0$ and it is not even a root.
[Note that over some fields, where the characteristic divides $n$, we do get repeated roots.]
Best Answer
As pointed out, the roots of
$t^n - 1 = 0 \tag 1$
are the $n$ complex numbers
$\omega^j = e^{2\pi i j / n} = (e^{2\pi i / n})^j, \; 0 \le j \le n - 1. \tag 2$
If we use the Euler identity on the $\omega^j$ we find
$\omega^j = e^{2\pi i j / n} = \cos \dfrac{2\pi j}{n} + i \sin \dfrac{2\pi j}{n}; \tag 3$
it is easy to see from $(3)$ that the ray emanating from the origin and passing through $\omega^j$ makes an angle $2\pi j / n$ with the positive $x$-axis; thus the angle between consecutive roots of unity $\omega^j$ and $\omega^{j + 1}$ is precisely $2\pi /n,$ no matter what the value of $j$; it is the same for any two consecutive $n$-th roots of unity, so the arc subtended by the angle 'twixt two consecutive $\omega^j$ always is of length $2 \pi / n$; they are evenly space around the unit circle.