prove that if p congruent to 3 (mod 4) is a prime and g is a primitive root mod p, then p - g is not a primitive root mod p.
p-g must be congruent to g^m mod p for some m, and if i want to show p - g is not a primitive root, then gcd(m,p-1) must not be 1, but I don't know how to proceed
Note that $g^{(p-1)/2} \ne 1 \mod p$ since $g$ is a primitive root, but since $(g^{(p-1)/2})^2 = g^{p-1} = 1 \mod p$ we must have $g^{(p-1)/2} = -1 \mod p$.
Now we can check directly that $p-g = -1\cdot g\mod{p}$ is not a primitive root. We have $(p-g)^{(p-1)/2} = (-1)^{(p-1)/2}g^{(p-1)/2} = (-1)(-1) = 1 \mod p$. Here we use the fact that $(p-1)/2$ is odd since $p = 3\mod 4$.
Another way of framing this is that your hypotheses imply that neither $g$ nor $-1$ are quadratic residues (ie, there is no value squaring to them), and the produce of two quadratic non-residues will always be a quadratic residue.