Do you know the theorem that says that if the orders of $c$ and $d$ are relatively prime, then the order of $cd$ is the product of the orders of $c$ and $d$?
The above result takes care of your problem. For suppose that $-a$ has order $(p-1)/2$. Since $p$ is of the form $4k+3$, $(p-1)/2$ is odd.
Also, $-1$ has order $2$. Thus the order of $(-1)(-a)$ is $p-1$.
Appendix: We prove the useful result about the order of a product. Suppose that $h$ is the order of $c$, and $k$ is the order of $d$.
Let $r$ be the order of $cd$. Clearly $(cd)^{hk}=(c^h)^k (d^k)^h \equiv 1 \pmod{p}$ and hence $r\mid hk$.
Also, $d^{rh}\equiv (c^h)^rd^{rh}\equiv (cd)^{rh}\equiv 1 \pmod{p}$ and hence $k \mid rh$. Since $(h,k)=1$, it follows that $k \mid r$. Similarly, $h \mid r$, and therefore $hk \mid r$, since $(h,k)=1$.
Since $r \mid hk$ and $k \mid r$, we conclude that $r=hk$.
Added: We only did one direction, since OP wrote that he had done the other direction. But for completeness, and in response to a request, we write out the part OP had no problem with. Let $p$ be a prime of the form $4k+3$, and suppose that $a$ is a primitive root of $p$. We show that $-a$ has order $(p-1)/2$.
Because $a$ is a primitive root, it is a quadratic non-residue. But $p$ has shape $4k+3$, so $-1$ is a quadratic non-residue. So $(-1)a$ is a quadratic residue. It follows that the order of $-a$ cannot be $p-1$. If the order of $-a$ is $e$, we have $(-a)^{2e}=a^{2e}\equiv 1\pmod{p}$. Thus $p-1$ divides $2e$, and therefore $e=(p-1)/2$.