(a) $g(0)=\int_0^0f(t)dt=0$
(b) g'(1)=\frac{d}{dx}(\int^x_0f(t)dt)\Big|_{x=1}=f(1)=-19 (0r -20? Or 19.5? I need to have a magnifying glass to see the exact value from the graph. Anyway, it's $f(1)$.)
(c) Note that \displaystyle\frac{d^2g}{dx^2}(x)=\frac{d^2}{dx^2}(\int^x_0f(t)dt)=f'(x). Therefore, $\displaystyle\frac{d^2g}{dx^2}(x)>0$ if and only if f'(x). That is, the interval where $g$ is convex is the same as the interval where $f$ is increasing, that is $(1,6)$ from the graph. (Again a magnifying glass is needed)
(d) Again, $\displaystyle\frac{dg}{dx}(x)=f(x)$. That is, the critical point of $g$ correspond to the zero of $f$. Therefore, $2$ is critical point of $g$, and $g(2)=\int_0^2f(t)dt<0$ since $f<0$ on $(0,2)$. Finally, to find the absolute maximum of $g$ on $[0,8]$, we need to check the boundary point: $g(0)=0$ by part (a), and $g(8)>0$. Therefore, $g$ takes its maximum at $x=8$.