= Solution
{c}
First consider the preliminary cohomological facts. The exam's term flabby means <flasque sheaf>: every restriction map is surjective. For open sets $V\subseteq U$, lift a section of $\mathcal F_3(V)$ to $\mathcal F_2(V)$ using the given surjectivity when $\mathcal F_1$ is flasque. Extend that lift to $U$ using the <flasque sheaf> $\mathcal F_2$, then project to $\mathcal F_3(U)$. This proves that $\mathcal F_3$ is a <flasque sheaf>.
One construction of a <flasque resolution> starts with the sheaf $\mathcal C^0(\mathcal F)(U)=\prod_{P\in U}\mathcal F_P$, with pointwise restriction. It is a <sheaf of abelian groups> with surjective restrictions, and sending a section to all its germs embeds $\mathcal F$ in it. Repeating on the cokernels constructs an <exact sequence>
$$
0\longrightarrow\mathcal F\longrightarrow\mathcal I^0\longrightarrow\mathcal I^1\longrightarrow\cdots
$$
with each $\mathcal I^j$ flasque. Define <sheaf cohomology> by the cohomology of the <cochain complex> of global sections of such a resolution; the groups are independent of the chosen <flasque resolution>. If $\mathcal F$ is itself flasque, the quotient result just proved makes every successive cokernel flasque. Applying the assumed surjectivity of sections to each successive <short exact sequence of sheaves> proves exactness of the global-section complex in positive degrees. Thus $H^i(X,\mathcal F)=0$ for $i>0$.
Now let $K=k(X)$ be the <function field> of the <smooth algebraic curve>. Its <divisor class group> is
$$
\operatorname{Cl}(X)=\operatorname{Div}(X)/\operatorname{Prin}(X),\qquad
\operatorname{Div}(X)=\bigoplus_{P\text{ closed}}\mathbb ZP,\qquad
\operatorname{div}(f)=\sum_P\nu_P(f)P\quad(f\in K^*).
$$
Here $\nu_P$ is the <discrete valuation> of the <discrete valuation ring> $\mathcal O_{X,P}$, and $\operatorname{Prin}(X)$ consists of the <principal divisor> elements $\operatorname{div}(f)$. These sums have finite support: on a finite affine cover, represent $f$ as a fraction of regular functions; each nonzero numerator or denominator has only finitely many zeros on a curve, since a proper closed subset of a Noetherian one-dimensional irreducible space is finite.
For a <divisor on an algebraic curve> $D=\sum_P n_PP$, define the <line bundle associated to a divisor> by
$$
\mathcal O_X(D)(U)=\{f\in K:\nu_P(f)+n_P\geq0\text{ for all closed }P\in U\}\cup\{0\}.
$$
In this expression the valuation condition applies to nonzero $f$. If $t_P$ is a <uniformizer>, the stalk is $t_P^{-n_P}\mathcal O_{X,P}$. Shrinking around $P$ removes all other zeros and poles of $t_P$ and all other points in the support of $D$, so this also gives an actual local generator. Thus $\mathcal O_X(D)$ is a <locally free sheaf> of rank one. Multiplication gives an isomorphism
$$
\mathcal O_X(D)\otimes_{\mathcal O_X}\mathcal O_X(E)\cong\mathcal O_X(D+E)
$$
on every stalk. If $D=\operatorname{div}(g)$, then $\mathcal O_X(D)=g^{-1}\mathcal O_X$, so the construction factors through a homomorphism $\operatorname{Cl}(X)\to\operatorname{Pic}(X)$.
It is injective: if $\mathcal O_X(D)$ is trivial, an isomorphism from $\mathcal O_X$ supplies a nonzero global rational generator $h$. At each closed point, $h$ generates $t_P^{-n_P}\mathcal O_{X,P}$, giving $\nu_P(h)=-n_P$, hence $D=-\operatorname{div}(h)$.
It is surjective: take a <line bundle> $\mathcal L$ and a nonzero rational section $s$, obtained by choosing a frame on any nonempty trivializing open set. If $e_i$ is a local frame, write $s=f_i e_i$ with $f_i\in K^*$. Since changes of frame are units, $\nu_P(f_i)$ is independent of the frame near $P$. A finite trivializing cover shows that these numbers have finite support. Set $D=\sum_P\nu_P(f_i)P$. The map $f\mapsto fs$ gives $\mathcal O_X(D)\cong\mathcal L$, because at $P$ the condition $\nu_P(f)+\nu_P(f_i)\geq0$ is exactly regularity of $fs$. Replacing $s$ by $gs$ changes $D$ by $\operatorname{div}(g)$, so this construction is well defined on classes. This proves the <divisor class group and Picard group of a smooth curve> identification
$$
\boxed{\operatorname{Cl}(X)\cong\operatorname{Pic}(X),\qquad[D]\longmapsto[\mathcal O_X(D)].}
$$
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