The value of a Dirichlet $L$-function (and its derivative) at $s=0$

This problem is missing some stuff. To get that kind of equality you need $\chi(-1)=-1$, since if $\chi(-1)=1$, then $L(0,\chi)=0$. I believe the question should be something like this:

The question: If $\chi(-1)=-1$, prove that $$L(0,\chi)=\frac{-1}{G(\chi)\sqrt{q}} \sum_{a=1}^q \chi(a)a,$$ where $G(\chi)$ is the Gauss sum.

Hint: So how can you do this? Start by getting an expression for $L(1,\chi)$ by switching sums in the definition of $L(s,\chi)$ with the identity $$\chi(n)=\frac{1}{G\left(\overline{\chi}\right)}\sum_{a=1}^{q}\overline{\chi}(a)e\left(\frac{an}{q}\right),$$ where $e(x)=e^{2\pi i x}$. Be careful about justifying convergence. It takes some work, and you have to argue why a certain sum with the logarithm of the sin function will cancel out. (This requires $\chi(-1)=-1$, and using the symmetry) Then apply the functional equation for $L(s,\chi)$ to get $L(0,\chi)$ from $L(1,\chi)$. (The $\kappa$ in the $\sin(x)$ factor is what causes $L(0,\chi)=0$ when $\chi(-1)=1$)

I hope that helps.