Answer
$(a)$
Local maximum: $(1,1)$
local minimum: $(2,0).$
$(b)$
Absolute maximum: $(1,1)$.
Absolute minimum: none.
$(c)$
See graph.
Work Step by Step
$f(x)=2x-x^{2}$
$f'(x)=2-2x=2(1-x)$
$f$ is defined on $(-\infty,2]$
$f'$ is defined on $(-\infty,2)$
$f'(x)=0$ for $ x=1\qquad$ ... critical point.
$f'(0)=2\gt 0$
$f'(2)=-2\lt 0$
$f':\quad \begin{array}{llllll}
-\infty & & 1 & & 2 & \\
( & ++ & | & -- & ] & \\\hline\\
& & 1 & & & \\
& \nearrow & & \searrow & 0 & \\
& & & & &
\end{array}$
