Answer
(a)
Increasing on $(0,4)$.
Decreasing on $(-\infty,0)\cup (4,5)$
(b)
No absolute maximum.
Local maximum: $(4,16)$
Absolute minima: $(0,0)$ and $(5,0).$
No other local minima.
Work Step by Step
$g$ is defined on $(-\infty,5]$.
$g'(x)=2x(5-x)^{1/2}+x^{2}(\displaystyle \frac{1}{2})(5-x)^{-1/2}(-1)$
$g'(x)=\displaystyle \frac{5x(4-x)}{2\sqrt{5-x}}$
which is not defined at $ x=5\qquad$ ... critical point.
$g'(0)=0$ for $x=0,4 \quad$... critical points.
$g(0)=0,\qquad g(4)=16,\qquad g(5)=0$
Using testpoints in the intervals between critical points,
$g'(-4)=-26.667... \lt 0$
$g'(1)=3.1 \gt 0$
$g'(4.5)=-0.795... \lt 0$
Create a table using the above:
$\begin{array}{l}
g':\\
\\
\\
g:
\end{array} \begin{array}{lllllll}
-\infty & & 0 & & 4 & & 5\\
( & -- & & ++ & | & -- & ]\\
& & & & & & \\
0 & \searrow & & \nearrow & 16 & \searrow & \\
& & 0 & & & & 0\\
& & & & & &
\end{array}$
