Answer
$(2,3)$.
Work Step by Step
Let's compare $f(x)=-2x^2+8x-5$ to $f(x)=ax^2+bx+c$. We can see that a=-2, b=8, c=-5. $a\lt0$, hence the graph opens down, hence its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{8}{2\cdot(-2)}=2.$ Hence the maximum value is $f(2)=-2(2)^2+8(2)-5=3.$ Hence the vertex is at $(2,3)$.
