The answer to Question 2 is yes. Here is an example of a tensor-commutative abelian group which is not a subquotient of $\mathbb Q$: $ M:=\mathbb Q\oplus\mathbb Q/\mathbb Z. $ Indeed, $M$ is tensor-commutative because the morphism mapping $(a,b)\otimes(c,d)$ (with $(a,b),(c,d)\in M$) to $ac$ is an isomorphism from $M\otimes M$ onto $\mathbb Q$, and $M$ is not a subquotient $S$ of $\mathbb Q$ because any such $S$ is torsion or torsion-free (more precisely, $S$ is a submodule of $\mathbb Q$ or a subquotient of $\mathbb Q/\mathbb Z$).
EDIT. The answer to Question 3 is no. Here is a proof.
Let $A$ be a tensor-commutative commutative ring, and let us show that $A$ is a subquotient of $\mathbb Q$.
We'll freely use Mariano's answer.
Assume $A\neq0$. Let $R$ be the prime ring of $A$.
Case 1: $R\simeq\mathbb Z/(n),\ n\ge2$. Claim: $A=R$.
We can assume that $n$ is a power of a prime. Then $A$ is an $R$-algebra containing $R$.
Suppose by contradiction that there is an $a$ in $A$ which is not in $R$. For any subgroup $G$ of $A$ containing $1$ and $a$, form the commutator $ c_G:=1\otimes a-a\otimes1\in G\otimes_RG=G\otimes_{\mathbb Z}G. $ It suffices to show $c_A\neq0$. It is even enough to check that $c_G$ vanishes for every finitely generated subgroup $G$ of $A$ containing $1$ and $a$. But this is clear because $R$ is a direct summand of $G$.
[Details: $G=R\oplus H;\ 1=(1,0);\ a=(r,h);\ h\neq0$.]
Case 2: $R=\mathbb Z$. Claim: $A\subset\mathbb Q$.
As explained by Mariano, it suffices to prove that $A$ is torsion-free.
The rest of the argument is very similar to the previous one:
Suppose by contradiction that there is a nonzero torsion element $t$ in $A$. For any subgroup $G$ of $A$ containing $1$ and $t$, form the commutator $ c_G:=1\otimes t-t\otimes1\in G\otimes_\mathbb ZG. $ It suffices to show $c_A\neq0$. It is even enough to check that $c_G$ vanishes for any finitely generated subgroup $G$ of $A$ containing $1$ and $t$.
Let $G$ be such a subgroup. Then $G$ is isomorphic to $\mathbb Z\oplus T$, where $T$ is the torsion subgroup of $G$, and the result is clear.
[Details: $G=\mathbb Z\oplus T;\ 1=(z,t_1);\ z\neq0;\ 0\neq t=(0,t)$.]