Answer
Neither, Even. See explanations.
Work Step by Step
Case 1. Assume $f(x)$ is odd: we have $f(-x)=-f(x)$, thus $g(-x)=f(-x)-2=-f(x)-2=-(g(x)+2))-2=-g(x)-4$. If $g(x)$ is odd, $g(-x)=-g(x)$ which leads to $-g(x)=-g(x)-4$ or $0=-4$ which is not possible. If $g(x)$ is even, $g(-x)=g(x)$ which leads to $g(x)=-g(x)-4$ or $g(x)=-2$ and $f(x)=0$ which means that both functions are constants. Thus the conclusion for this case is that except for $f(x)=0, g(x)=-2$, function $g(x)$ will be neither even nor odd.
Case 2. Assume $f(x)$ is even: we have $f(-x)=f(x)$, thus $g(-x)=f(-x)-2=f(x)-2=g(x)$ which means that function $g(x)$ will also be even.
