#### Answer

$g$ is discontinuous at $x=1$; it is left-continuous there.
$g$ is discontinuous at $x=3 ;$ it is neither left-continuous, nor right-continuous there.
$g$ is discontinuous at $x=5$; it is right-continuous there.
We have a removable discontinuity at $x=3$.

#### Work Step by Step

From the given figure, we have
$$\lim_{x\to 1^-}g(x)=g(1)=2$$
and\begin{align*}
\lim_{x\to 3^-}g(x)&=4.5 \\
\lim_{x\to 3^+}g(x)&=4.5\\
g(3)&=2.5
\end{align*}
and \begin{align*}
\lim_{x\to 5^-}g(x)&=3.5 \\
\lim_{x\to 5^+}g(x)&=1\\
g(5)&=1
\end{align*}
Thus, we see that:
$g$ is discontinuous at $x=1$; it is left-continuous there.
$g$ is discontinuous at $x=3 ;$ it is neither left-continuous, nor right-continuous there.
$g$ is discontinuous at $x=5$; it is right-continuous there.
We have a removable discontinuity at $x=3$ because if we had $g(3)=4.5$ instead of $g(3)=2.5$, the function would be continuous there.