= Solution
The <uniformization theorem> states that every simply connected <Riemann surface> is <biholomorphic> to exactly one of the <Riemann sphere> $\widehat{\mathbb C}$, the <complex plane> $\mathbb C$, or the <unit disc> $\mathbb D$. A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by $\widehat{\mathbb C}$ is the sphere itself, because every nonidentity Möbius transformation has a fixed point and therefore cannot act as a deck transformation. The <Riemann surfaces uniformized by the complex plane> are
$$
\boxed{\mathbb C,\qquad\mathbb C^*,\qquad
\mathbb C/\Lambda\text{ for a lattice }\Lambda.}
$$
Let $U\subseteq\mathbb C$ have a complement containing at least two points. Its universal cover cannot be the sphere because $U$ is noncompact. If it were the plane, the covering map $\mathbb C\to U\hookrightarrow\mathbb C$ would be a nonconstant <entire function> omitting two values, contrary to the <Little Picard theorem>. The <plane domain with two omitted points is hyperbolic>, so
$$
\boxed{U\text{ is uniformized by }\mathbb D.}
$$
Now put $S=R\setminus\{P_1,\ldots,P_n\}$. The spherical possibility occurs only for $(g,n)=(0,0)$. By the preceding list, the plane-uniformized possibilities are
$$
(g,n)=(0,1),(0,2),(1,0),
$$
which give respectively $2g-2+n=-1,0,0$. Every remaining pair has $2g-2+n>0$ and must have disc universal cover. Thus the <uniformization of a punctured compact Riemann surface> gives
$$
\boxed{
\begin{aligned}
S\text{ is uniformized by }\mathbb D
&\Longleftrightarrow 2g-2+n>0,\\
S\text{ is uniformized by }\mathbb C
&\Longleftrightarrow 2g-2+n\in\{-1,0\}.
\end{aligned}}
$$
For the <complex torus> $X=\mathbb C/\Lambda$, any analytic map $f:\mathbb C\to X$ lifts through the <universal covering map> to an entire function $F:\mathbb C\to\mathbb C$. If $F$ is constant, so is $f$. Otherwise Little Picard says that $F$ omits at most one complex number. For every $x\in X$, its lifts form an infinite coset $z+\Lambda$, so at least one lift is attained by $F$. Hence the <holomorphic map from the complex plane to a one-dimensional complex torus> satisfies
$$
\boxed{f\text{ is constant or surjective}.}
$$
Finally, $\mathbb C$ and $\mathbb D$ are homeomorphic by
$$
z\longmapsto\frac{z}{1+|z|},
\qquad
w\longmapsto\frac{w}{1-|w|}.
$$
They are not <conformally equivalent>: a biholomorphism $\mathbb C\to\mathbb D$ would be a nonconstant bounded entire function, contradicting the <Liouville theorem>. This proves the claimed example of <complex plane and unit disc are homeomorphic but not conformally equivalent>.
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