How to proceed in Question 15?
First it is obvious $x=0$ will be a root. Then I put $x=1/x$ and saw that the equation will be same. So, if $z$ is a root then $1/z$ will also be a root. Also, the equation has rational coefficients, so imaginary roots will come in conjugate pairs. After that I am stuck, I am not able to proceed.
Let $c_k$ be the coefficients of the polynomial. If you make a plot of $(k,c_k)$, you will notice it forms a symmetric triangle. Recall if you convolute two step functions, you get a symmetric triangle. This means the polynomial can be factored as
$$x(1+x+x^2+\cdots+x^{23})^2 = x\left(\frac{x^{24}-1}{x-1}\right)^2$$
As a result, the non-zero distinct roots of it has the form $\omega^k, k = 1,\cdots, 23$ where $\omega = e^{\frac{\pi}{12}i}$ is the primitive $24^{th}$ root of unity. The sum you want is simply $\sum_{k=1}^{23}\left|\Im \omega^{2k}\right|$.
Since $\Im(\omega^{2(k+6)}) = -\Im \omega^{2k}$, we have $$\sum_{k=1}^{23}\left|\Im \omega^{2k}\right| = \sum_{k=0}^{23}\left|\Im \omega^{2k}\right| = 4 \sum_{k=0}^{5}\left|\Im \omega^{2k}\right| = 4\left(\frac12 + \frac{\sqrt{3}}{2} + 1 + \frac{\sqrt{3}}{2} + \frac12\right) = 8 + 4\sqrt{3}$$