Using LES:
$$
H_2(X)\to H_2(X,A)\to H_1(A)\to H_1(X)\to H_1(X,A)\to H_0(A)\to H_0(X)\ldots
$$
As $H_2(X)=0$, the second arrow is an injection. The third arrow, after the obvious identifications, is $(a,b)\mapsto b$ which has kernel $\{(a,0)\}$
and hence $H_2(X,A)\simeq \Bbb Z$. It is generated by the meridian $2$-cell that has boundary in $A$. The last map is an isomorphism, so the second-last is a zero map; but the third map is onto, so the fourth map is zero as well: it follows that $H_1(X,A)=0$. (Intuitively, the only candidate for $H_1(X,A)$ is a circle going around the solid torus, but this can be homotoped to the boundary).
No, the algebraic Kunneth sequence tells us that if we have free chain complexes C,D over the PID R, then we have the short exact sequence:
$0 \rightarrow \bigoplus\limits_{i+j=k} H_i(C) \otimes_R H_j(D) \rightarrow H_k (C \otimes D) \rightarrow \bigoplus\limits_{i+j=k-1} Tor^R_1 (H_i(C),H_j(D)) \rightarrow 0$.
The Eilenberg-Zilber theorem tells us that $S_*(X \times Y,R)$ and $S_*(X,R) \otimes S_*(Y,R)$ are quasi-isomorphic. From these one can deduce the usual Kunneth formula in topology. To figure out the case for coefficients in a module M we can just tensor each of these free chain complexes by $M$ and we preserve the quasi-isomorphism. So we have $S_*(X \times Y,R) \otimes_R M$ is quasi-isomorphic to $S_*(X,R) \otimes S_*(Y,R) \otimes_R M$. Rewriting, the former is $S_*(X \times Y,M)$ and the latter is $S_*(X,R) \otimes_R S_*(Y,M)$.
Applying the algebraic Kunneth sequence tells us that we have a short exact sequence:
$0 \rightarrow \bigoplus\limits_{i+j=k} H_i(X,R) \otimes_R H_j(Y,M) \rightarrow H_k (X \times Y,M) \rightarrow \bigoplus\limits_{i+j=k-1} Tor^R_1 (H_i(X,R),H_j(Y,M)) \rightarrow 0$
And of course you can swap the roles of X and Y if you'd like to pick which space gets which coefficients. The reason why your proposed sequence does not work is because there are modules such that $M \otimes_R M \neq M$.
In the case R is not a PID, you can use the algebraic Kunneth spectral sequence because I believe the Eilenberg-Zilber theorem holds with no conditions on the ring. This spectral sequence involves the higher Tor terms which vanish for PID's.
Best Answer
Write the two torus as $S^1 \times S^1$. Use that $H_0(S^1) = H_1 (S^1) = \mathbb Z$ and $H_i (S^1) = 0$ for $i > 1$.
Also use the fact that $\mathbb Z \otimes_{\mathbb Z} \mathbb Z = \mathbb Z$.
The answers you would get would be $H_0(T^2) = \mathbb Z$.
$H_1(T^2) = \mathbb Z \oplus \mathbb Z$.
$H_2(T^2) = \mathbb Z$.
$H_i(T^2) = 0$ for $i >2$.
Now I must say that the Kunneth formula for homology has some $\text{Tor}$ terms. It works over here because homology groups of sphere are free.
That formula won't work for say $\mathbb RP^2 \times \mathbb RP^2$.
Anyway, that formula does work for coefficients in a field, since all modules over a field are free.
To relate product of three or more spaces, you can just apply the Kunneth formula repeatedly.
So for n torus, use the Kunneth formula repeatedly to obtain
$$H_k(S^1\times\dots\times S^1) = \bigoplus_{i_1 + \dots + i_r = k}H_{i_1}(S^1)\otimes\dots\otimes H_{i_r}(S^1).$$
Note that nonzero things appear only when the indices are $0$ or $1$. Use it to get the following :
$$H_k(T^n) = \mathbb Z ^{\binom{n}{k}}.$$