#### Answer

(a) We cannot conclude anything about the values of $f, g, h$ at $x=2$.
(b) $f(2)=0$ is possible.
(c) $\lim_{x\to2}f(x)=0$ is not possible.

#### Work Step by Step

$g(x)\le f(x)\le h(x)$ for all $x\ne2$
$\lim_{x\to2}g(x)=\lim_{x\to2}h(x)=-5$
According to the Sandwich Theorem, this means $\lim_{x\to2}f(x)=-5$
(a) Can we conclude anything about the values of $f, g, h$ at $x=2$?
Unfortunately, no. Recall that when we are given or need to find the limit of a function $f(x)$ as $x\to c$ (or $x\to2$ in this case), we measure the value that $f(x)$ would reach as $x$ APPROACHES $c$.
So we do not actually work with the value of $f(x)$ as $x=c$; hence, it is not necessarily true that $f(c)=\lim_{x\to c}f(x)$ (a lot of cases have been shown to be wrong in this section).
Therefore, we cannot conclude anything about the values of $f, g, h$ at $x=2$. They might equal $-5$, or they might not.
(b) Could $f(2)=0$?
Yes, it is possible. Or $f(2)=1$, $f(2)=2$, $f(2)=-98/5$, etc. any of them would also be possible.
(c) Could $\lim_{x\to2}f(x)=0$?
No. As proved according to the Sandwich Theorem above, $\lim_{x\to2}f(x)=-5$
Notice that the inequality $g(x)\le f(x)\le h(x)$ does not have to be true for $x=2$, for reasons, again, that we do not work with $x=2$ when we work with limit as $x\to2$.