I don't know about 2., but the answer to 1. is yes.
More generally, suppose $(S^1)^g\times\mathbb Z^k \cong (S^1)^h \times \mathbb Z^m$ as abstract groups, then $g=h, k=m$ (in fact, you only need $g=h$ for question 1., because the abstract isomorphism type of $\mathbb C^g/\Lambda$ only depends on $g$, and because you already know $k=m=2$)
(Note that this gives part of the answer for 2., namely that $C_1$ and $C_2$ have the same genus)
Let's prove this statement.
First, if there is an isomorphism between the two groups, then their torsion parts are isomorphic. Their torsion parts are $(\mathbb{Q/Z})^g$ and $(\mathbb{Q/Z})^h$ respectively. But you can recover $g$ from $(\mathbb{Q/Z})^g$ in the following way : take the $2$-torsion (not $2$-power torsion, $2$-torsion) part as an $\mathbb F_2$-vector space, this has dimension $g$. (I chose $2$ but could have chosen any $p$, in fact for any integer $n\geq 2$, the $n$-torsion part is a free $\mathbb Z/n$-module of rank $g$, and these rings have the invariant basis property).
Therefore $g=h$ - this is all that's relevant to your question, but let me still prove $k=m$.
Note that to get $k=m$, we cannot rationalize and take dimensions because $(S^1)^g\otimes \mathbb Q$ is an uncounatbly dimensional rational vector space.
We now observe that we can mod out the divisible part - I don't know the standard name, but let me define it : $\mathrm{Div}(A) = \{a\in A\mid \forall n \neq 0, \exists b \in A, nb = a\}$. This is clearly a subgroup of $A$, and $\mathrm{Div}(A\times B)\cong \mathrm{Div}(A)\times \mathrm{Div}(B)$. In particular $\mathrm{Div}((S^1)^g\times \mathbb Z^k) = (S^1)^g\times \{0\}$, and so modding out $A/\mathrm{Div}(A)$ gives, in our case, an isomorphism $\mathbb Z^k\cong \mathbb Z^m$, and therefore $k=m$.
Best Answer
If your question concerns - as mentioned in one of your comments - if there is any relationship between them, then a very beautiful connection exists in what is called geometric class field theory:
namely classical number theoretic class field theory concentrates around what is called Artin Reciprocity, which establishes an isomorphism for a number field $K$ and its ring of integers $\mathcal{O}_{K}$ an isomorphism $Pic(Spec(\mathcal{O}_{K})) \cong \pi_{1}^{ab}(Spec(\mathcal{O}_{K}))$ between the Picard group and the abelianized etale fundamental group (it is a geometric reformulation of classical Artin reciprocity). We can see it as a special case of one-dimensional class field theory and the question arises naturally if we can extend somehow this correspondence for higher dimensions (and also for other one dimensional schemes). There are different approaches (K-theory, cycle theory, geometric Langlands) but the main cornerstones are the following:
Bloch-Kaito-Saito Theorem: Let $X$ be a regular, connected, projective scheme over $Spec(\mathbb{Z})$, then there exists also a reciprocity map $Pic(X) \rightarrow \pi_{1}^{ab}(X)$ which is an isomorphism if in addition $X$ is flat over $Spec(\mathbb{Z})$. If $X$ factors through a finite field $k=\mathbb{F}_{q}$ then the reciprocity map is injective and with cokernel $\widehat{\mathbb{Z}}/\mathbb{Z}$.
Also for curves over finite fields there exists a correspondence, namely if $C$ is a smooth, projective, geometrically irreducible curve over a finite field $k$, then there exists a reciprocity homomorphism $Pic_{C}(k) \rightarrow \pi_{1}^{ab}(C)$ which induces an isomorphism on the degree zero parts $Pic_{C}^{0}(k) \rightarrow \pi_{1}^{ab}(C)^{0}$, where the degree maps are the obvious maps to $\mathbb{Z}$ and $\widehat{\mathbb{Z}}$ resp.
Also if $S \subset C$ is a finite set of points of a smooth, projective, geom. irreducible curve $C$ over a finite field, then there is a ramified version of the previous reciprocity, namely between $Pic_{C,S}$ (which is isomorphism classes of line bundles together with fixed isomorphisms of the stalks at every point in $S$) and the abelianization of the tame fundamental group of $U:=C \setminus S$.
Some reference: http://epub.uni-regensburg.de/13979/1/MP92.pdf
and then it gives many other references and so on...