uncountable ring with countable quotient ring?

Let $R$ be any finite or countably infinite commutative ring except the one ring. Then $R[[x]]$, the ring of formal power series in the variable $x$ over $R$ is uncountable and $R[[x]]/(x)$ is isomorphic to $R$, so is finite or countably infinite.

Yes. For instance, for any rings $A$ and $B$, the projection $A\times B\to B$ gives an isomorphism between $B$ and a quotient ring of $A\times B$. In particular, if $A$ is uncountable then $A\times B$ will be uncountable, but $B$ could be any ring at all. Or, for any ring $B$, you can form a polynomial ring $R$ over $B$ in uncountably many variables, which will be uncountable (assuming $B$ is nonzero) but has a quotient isomorphic to $B$ (just mod out the ideal generated by all the variables).

A more naturally occurring example is the ring $\mathbb{Z}_p$ of $p$-adic integers, which is uncountable, but the quotient $\mathbb{Z}_p/(p)\cong \mathbb{F}_p$ is finite.

I believe $R=\Bbb Z\times\Bbb C$ and $I=\{0\}\times\Bbb C$ works.