Answer
$ a.\quad$ No
$ b.\quad$ Only one extreme value of f, and it occurs at x=2.
$ c.\quad$ No contradiction.
$ d.\quad f'(a)$ is not defined. $x=a$ is the only critical point, so f has only one extreme value, and it occurs at $x=a.$
Work Step by Step
$f'(x)=\displaystyle \frac{2}{3}(x-2)^{-1/3}\cdot(1)=\frac{2}{3(x-2)^{1/3}}$,
$x=2$ is the only critical point, because
$f'(x)$ is undefined at x=2, and positive for all other x.
$ a.\quad f'(2)$ does not exist
$ b.\quad$
At x=2, $f(x)=0$, and for all other x, $f(x)$ is positive.
There is only one critical point, so the only extreme value occurs at x=2.
$ c.\quad$ The Extreme value theorem refers to a function whose domain is a closed interval, and this is not the case here. No contradictions.
$ d.\quad$
Replacing 2 with a, sections (a) and (b) have the same answers.
