Tyler, thanks for posting a question.

You said:

I have done out the problem in a sense by plugging in 0 for x. I found that h(0)= -3 because 2(0)-3= -3

This is a huge part of what you need to do, but there is a little bit more to it. In order to address the "Show that it is continuous at the point $x=0$" part of the question, you need verify that

(1)
\begin{align} \lim_{x\to 0}h(x)=-3=h(0). \end{align}

The loosey-goosey way of saying this is that you need to check that "near"="at" (or that nothing weird happens at $x=0$).

You say:

I believe that it's not continuous at the point x=0…

I don't think that this is right…actually, we know it isn't since the problem asked us to show that it is continuous at $x=0$ (and I know this is correct). Again, to prove this, verify the identity above (which pretty much amounts to just writing it down since "near" 0 nothing weird happens on $2x-3$).

To address the "Is $h$ a continuous function?" part of the exercise, you need to determine whether something weird is happening somewhere. If the function isn't continuous, then something weird must be happening at $x=1$ since each piece of the piecewise function is individually continuous. The question boils down to asking whether the two pieces meet up at $x=1$.

I've intentionally not given everything away. However, if you are still struggling with understanding, ask some more questions.