How can I prove that if $\chi$ is a non-principal character modulo $p$ prime, then $\chi (-1) = \overline{\chi} (-1)= \pm 1$ and $\sum_{x=1}^p \chi (x) e^{2\pi i x}=0$?
For the first question, I just know that $\chi(-1)\overline{\chi} (-1)=1$, but they could be complex?
And for the second question: no idea...
We have $1=\chi(1)=\chi((-1)^2))=\chi(-1)^2$. Since the square roots of $1$ are $1$ and $-1$, we obtain $\chi(-1)=\pm 1$. Also, $\overline{\chi (-1)}=\chi(-1)^{-1}=\pm 1$. For the second part, I show that another sum is zero, i.e., $S=\sum_t\chi(t)$. Note that there is an $a\neq 0$ such that $\chi(a)\neq 1$, because $\chi$ is non-principal. Let $S=\sum_t\chi(t)$. Then $$ \chi(a)S=\sum_t\chi(a)\chi(t)=\sum_t\chi(at)=S. $$ So we have $S(\chi(a)-1)=0$. Since $\chi(a)\neq 1$, it follows $S=0$. This is also used for proving that the Gauss sum $G(1,\chi)=\sum_t \chi(t)e^{\frac{2\pi it}{p}}$ has absolute value $\sqrt{p}$. Your sum is similar.