4. By the fundamental theorem of calculus,
[tex]\displaystyle\frac{\mathrm dg(x)}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\int_{-2}^xf(t)\,\mathrm dt=f(x)[/tex]
[tex]g(x)[/tex] is increasing on those intervals where [tex]g'(x)=\dfrac{\mathrm dg(x)}{\mathrm dx}>0[/tex]. So you have
[tex]g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x<0\\-x+3&\text{for }0\le x\le6\\-3&\text{for }6<x\le9\end{cases}[/tex]
which is clearly positive for [tex]x\in[-3,0)[/tex], and in the second interval you have
[tex]-x+3>0\implies x<3[/tex]
Together, this means [tex]g'(x)>0[/tex] for all [tex]x\in[-3,3)[/tex].
5. When [tex]0\le x\le6[/tex], [tex]f(x)[/tex] reduces to [tex]-x+3[/tex], so you have
[tex]g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt[/tex]
[tex]g(x)=6+\left(3t-\dfrac12t^2\right)\bigg|_{t=0}^{t=x}[/tex]
[tex]g(x)=6+3x-\dfrac12x^2[/tex]
