I'm assuming that you want to find values of $p$ and $q$ so that the function $f$ is continuous at $x=0$.
To do this, you need $\tag{1}\lim_{x\rightarrow0^-}f(x) = f(0) =\lim_{x\rightarrow0^+}f(x).$
You know that $f(0)=q$. You might eventually set $\lim\limits_{x\rightarrow0^-}f(x)=q$ in the solution; but you need to find the values of $p$ and $q$ so that $(1)$ holds. And to do this, you need to find the one-sided limits first.
Where do you get, from your work, that $2+p=0$? I'm not sure what you are doing here...
So, again, what you should do is evaluate the one-sided limits in $(1)$ first. You need to find
$\ \ \ \lim\limits_{x\rightarrow0^-} f(x)=\lim\limits_{x\rightarrow0^-} {2x+p\sin x\over 3x-2\sin x} $
and
$\ \ \ \lim\limits_{x\rightarrow0^+} f(x)=\lim\limits_{x\rightarrow0^+} {3+2x\cot x\over 2+x\csc x} $.
From your answer key, you have $ \lim_{x\rightarrow0^-}f(x) =2+p,\quad \text{and}\quad \lim_{x\rightarrow0^+}f(x)={5/3}. $ Now let's see how to obtain $(1)$. The limit from the right is $5/3$ and $f(0)=q$, so $q$ is $5/3$ and that gives the equality on the right hand side of $(1)$. To get the equality on the left hand side, we set $5/3$ equal to the limit from the left: $5/3=2+p$, so $p=-1/3$.