When $I$ is generated by only one element I find it easy to write down the set $\mathbb{Z}[x]/I$ to find ideals ecc... However when there are more generator I'm not sure how it works.
I think that if $I=(a,b)=(a)+(b)=:A+B$ we have $$\mathbb{Z}[x]/(A+B) \cong (\mathbb{Z}[x]/A)/((A+B)/A) \cong (\mathbb{Z}[x]/A)/ \pi(B)$$
where $\pi$ is the canonical projection from $\mathbb{Z}[x]$ to $\mathbb{Z}[x]/A$, so $\pi(B)$ is $B$ viewed as an ideal of $\mathbb{Z}[x]/A$.
I know that $\mathbb{Z}[x]/(x,3) \cong \mathbb{Z}/3\mathbb{Z}$ and the proof i saw was $$\mathbb{Z}[x]/(x,3) \cong (\mathbb{Z}[x]/(3))/((x,3)/(3) \cong (\mathbb{Z}/3\mathbb{Z}[x])/(x) \cong \mathbb{Z}/3\mathbb{Z}$$
and here is the first question: why $(x,3)/(3) \cong (x)$? Is it because i have to view $(x)$ in $\mathbb{Z}/3\mathbb{Z}[x]$ so the $(x)$ above should be $\pi((x))$?
with more generator is it still the same? for example if $I=(5,3x,x^2)$ can I say that $$\mathbb{Z}[x]/I \cong \mathbb{Z}/5\mathbb{Z}[x]/J$$
where $J=(3x,x^2)$ ideal of $\mathbb{Z}/5\mathbb{Z}[x]$ and so
$$\mathbb{Z}/5\mathbb{Z}[x]/J \cong ((\mathbb{Z}/5\mathbb{Z}[x]/)/(3x))/(x^2)$$
where $(3x)$ is viewed in $\mathbb{Z}/5\mathbb{Z}[x]$ and $(x^2)$ is in $(\mathbb{Z}/5\mathbb{Z}[x]/)/(3x)$?
Is it correct or am I totally wrong? I don't think this make sense. I would then say that $$\mathbb{Z}[x]/(5,3x,x^2)= \{a_0+a_1x | a_0 \in \{0,1,2,3,4 \}, a_1 \in \{0,1,2 \}, x^2=0 \} $$
but I am pretty sure that this is wrong
On your first question, the natural map $\mathbb{Z}[x]/(3)\rightarrow\frac{\mathbb{Z}}{3\mathbb{Z}}[x]$ is given by $p+(3)\mapsto\bar{p}$. So an element $fx+3g\in(x,3)$ is send to $\bar{f}x+\bar{3}\bar{g}=\bar{f}x\in(x)$. On the other hand, if $\bar{h}x\in(x)$ where $h\in\mathbb{Z}[x]$ then $hx\in(x)\subset(x,3)$ is such that $hx\mapsto\bar{h}x$. Since its an isomorphism we must have $(x,3)/(3)\simeq(x)$.
The same happens with your other example. You need to have in mind the isomorphism theorem: if $J\subseteq I$ are ideals in $R$ then
$$\frac{R/J}{I/J}\simeq R/I.$$