This is response to QUESTION 1.
As Fedor pointed out, we're dealing with the Chebyshev polynomials $P_n(2\cos t)=\sin nt/\sin t$. So we must show that if $$ \sum_{n=1}^N \sin nt = 0 , \quad\quad\quad\quad (1) $$ then also $\sum_{n=1}^N \sin^m nt = 0$ for any odd exponent $m\ge 1$.
We may take $0<t<\pi/2$. Also, the sum in (1) can of course be evaluated, and we find that (1) is equivalent to $$ \cos t/2 = \cos (N+1/2) t . \quad\quad\quad\quad (2) $$ I now claim that if (2) holds for $t$, then it also holds for any multiple of $t$. To see this, we just notice that (2) means that $s=t/2$ satisfies $(2N+1)s = 2\pi M\pm s$, for some $M\ge 1$ and a choice of sign (recall that $0<s<\pi/4$). In other words, (2) requires $s$ to be a rational multiple of $\pi$ with denominator $N$ or $N+1$, and clearly this property is preserved under taking integer multiples.
Now everything is clear: $\sin^m\alpha$ can be written as a linear combination of $\sin j\alpha$, with $j$ odd, and we have just seen that (1) implies that also $\sum_{n=1}^N \sin njt = 0$ for any odd $j$.
