#### Answer

$\lim_{x\to-2^+}f(x)=\lim_{x\to2^-}f(x)=7$
$\lim_{x\to-2^-}f(x)$ cannot be decided here.

#### Work Step by Step

If $f$ is an even function of $x$, then $f(-x)=f(x)$.
We know that $\lim_{x\to2^-}f(x)=7$. This means as $x$ approaches $2$ from the left $(x\lt2)$, the values of $f(x)$ gets closer and closer to $7$.
$f$ is an even function, so its graph would be symmetric through $Oy$. All the values of $f(x)$ to the left of $2$ would be reflected in the values of $f(x)$ to the right of $-2$.
So if $x$ approaches $-2$ from the right, as $-x$, $f(-x)=f(x)$, the values of $f(x)$ would also approach $7$.
In other words, $\lim_{x\to-2^+}f(x)=\lim_{x\to2^-}f(x)=7$
The value of $\lim_{x\to-2^-}f(x)$, unfortunately, cannot be decided without the knowledge of $\lim_{x\to2^+}f(x)$.
A graph of an even function is shown below for clarification. As you can see, $\lim_{x\to-2^+}f(x)=\lim_{x\to2^-}f(x)=7$ here, but $\lim_{x\to-2^-}f(x)$ would not exist here because the graph is not defined in the domain $(-\infty,-2)$.