= Solution
First, $1\in G(p)$. Otherwise there would be a nonconstant <rational function> with one simple pole and no others, defining a degree-one <finite morphism> $X\to\mathbb P^1$. Such a map is an <isomorphism of algebraic varieties>[isomorphism], which would make $X$ have genus zero.
Part (iii) says that there are exactly two gaps. If $2$ is a gap, they must therefore be $1$ and $2$. If $2$ is not a gap, choose a function with its only pole at $p$, of order two. Pole orders form an <additive semigroup> under multiplication of functions. If $3$ were also a nongap, the semigroup generated by $2$ and $3$ would contain every integer at least two, leaving only the single gap $1$, a contradiction. Hence $3$ is the second gap. This proves the <Genus-two Weierstrass gap sequence>
$$
\boxed{G(p)=\{1,2\}\quad\text{or}\quad G(p)=\{1,3\}.}
$$
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