= Solution
A <Riemann surface> is a connected one-complex-dimensional manifold with <holomorphic> chart transitions. A map between such surfaces is <holomorphic> when all its local coordinate expressions are <holomorphic>; it is <biholomorphic> when it is a bijection whose inverse is also <holomorphic>.
The open mapping theorem says a nonconstant <holomorphic> function on a connected complex domain maps open sets to open sets. Applied in charts, with the <identity theorem> excluding a locally constant portion, it makes a nonconstant <holomorphic map> of connected surfaces open. If the domain is compact, its image is compact and hence closed in the Hausdorff target. A nonempty set both open and closed in a connected target is the whole target. Therefore the map is surjective, without any appeal to degree.
For a biholomorphism $\phi:\Delta^*\to\Delta^*$, both $\phi$ and its inverse are bounded near the missing origin. They extend holomorphically across zero by removability: in a <Laurent series> a negative coefficient satisfies $|a_{-m}|\leq M r^m$ on radius $r$, hence vanishes as $r\to0$. Let the extensions be $F,G$. Their images lie in $\Delta$: a modulus-one interior value would force a bounded <holomorphic map> to be constant by the <maximum modulus principle>. The compositions $GF$ and $FG$ equal the identity on the punctured disk and thus on the disk by <continuity>. If $F(0)\ne0$, then $G(F(0))$ would be nonzero because the original inverse maps $\Delta^*$ into itself, contradicting $G(F(0))=0$. Consequently
$$
\boxed{F:\Delta\to\Delta\text{ is biholomorphic and }F(0)=0.}
$$
For the final assertion only <continuity> on $R$ and holomorphicity on $R\setminus\{p\}$ are needed; the printed stronger description is redundant. Choose a target chart about $f(p)$ and, by <continuity>, a small source chart mapped into it. The resulting scalar function is <holomorphic> on a punctured disk and bounded near its center, so the same Laurent argument removes its singularity. <continuity> makes the extension's central value equal the prescribed value. Thus $f$ is <holomorphic> also at $p$.
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