This solution relies on the long exact reduced homology sequence of a NDR pair (Hatcher's Theorem $2.13$), and provides different approach to the problem from @tsho's solution.
Let us call $$\underbrace{\Huge{\mathsf x} \normalsize\times S^1}_{A}~~\subset~~
\underbrace{\Huge{\propto}\normalsize\times S^1}_{B}~~\subset~~
\underbrace{\Huge{\infty}\normalsize \times S^1}_{X}$$ All three pairs $(B,A),~(X,A),~(X,B)$ are Neighborhood Deformation Retracts, as Hatcher puts it, "good pairs". Also, it is obvious that $A$ is homotopy equivalent to the circle, and $B$ to the torus. Let us write the long exact reduced homology sequence for the the good pairs $(X,A)$ and $(X,B)$ : the morphism of pair given by the inclusion $(X,A)\hookrightarrow (X,B)$ gives following commutative diagram
$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/A)&\to&\tilde{H}_1(A)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/A)&\to 0\\
&\downarrow&&\Vert&&\downarrow&&\downarrow&&\Vert&&\downarrow\\
0\to& \tilde{H}_2(B)&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/B)&\to&\tilde{H}_1(B)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/B)&\to 0
\end{array}$$
Now $X/A$ is the wedge sum of two pinched spheres $P$ (the space studied in the previous question), and $X/B$ is simply a pinched sphere, and it follows that $\tilde{H}_*(X/B)\simeq\tilde H_*(P)$ and $\tilde{H}_*(X/A)\simeq\tilde H_*(P)\bigoplus \tilde H_*(P)$ where the isomorphism is given by the map $(i_*^+,i_*^-)$ where $i^+$ (resp. $i^-$) are the inclusions of $P$ as the upper (resp. lower) pinched sphere in $X/A$.
Since a pinched sphere is homotopy equivalent to a sphere with a diameter attached to it, which in turn is homotopy equivalent to the wedge sum of a sphere and a circle, we have $\tilde H_*(P)\simeq \Bbb Z\oplus\Bbb Z$ concentrated in degree $1$ and $2$. We can now replace the above diagram by the following simpler one
$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\Bbb Z\oplus \Bbb Z &\stackrel{\gamma}{\to}&\Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z\oplus \Bbb Z &\to 0\\
&\downarrow&&\Vert&&~~\downarrow\beta&&\downarrow&&\Vert&&\downarrow\\
0\to& \Bbb Z &\to&\tilde{H}_2(X)&\stackrel{\alpha}{\to}& \Bbb Z &\to& \Bbb Z\oplus \Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z &\to 0
\end{array}$$
From the left side of this diagram, it follows that $\tilde H_2(X)$ is a subgroup of $\Bbb Z\oplus \Bbb Z$ containing a copy of $\Bbb Z$, so $\tilde H_2(X)\simeq\Bbb Z$ or $\Bbb Z\oplus\Bbb Z$. Let us assume $\tilde H_2(X)\simeq \Bbb Z$ and try to derive a contradiction.
Since $\Bbb Z$ is torsion free, we must have $\alpha=0$. The vertical map $\beta$ is onto as it corresponds to collapsing the lower copy of $P$ inside $P\vee P\simeq X/A$ to a point, and thus $\beta$ is the projection onto the first factor. The commutativity of the diagram then forces the image of $\tilde H_2(X)$ to lie inside $\Bbb Z\oplus 0\subset \Bbb Z\oplus \Bbb Z$. However, there is an obvious self-homeomorphism of $X$ interchanging the upper and lower toruses of $X$ which passes to the quotient, and permutes the two factors $\Bbb Z \oplus \Bbb Z=\tilde H_2(X/A)$ (and possibly adds a sign). Thus, the image of $\tilde H_2(X)$ inside $\Bbb Z \oplus \Bbb Z$ is contained in $\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$, but this contradicts the injectivity of the top left arrow. The same argument works when we replace $B$ with $B'=T(B)$ where $T$ is the self-homeomorphism of $X$ that interchanges the two circles in the wedge sum $S^1\vee S^1$. The new map $\beta'$ is the projection onto the second factor, so the map
$\tilde H_2(X)\to \Bbb Z\oplus\Bbb Z$ sends $\tilde H_2(X)$ into $\ker(\beta)\cap\ker(\beta')=\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$ contradicting injectivity.
As a consequence, $$\tilde H_2(X)\simeq \Bbb Z\oplus\Bbb Z$$
To finish the proof, we note that by the standard theory of finitely generated abelian groups, the quotient of $\Bbb Z\oplus\Bbb Z$ by a subgroup $S$ isomorphic to $\Bbb Z\oplus\Bbb Z$ is the product of two cyclic groups, and cannot be a subgroup of $\Bbb Z$ unless the subgroup $S$ is all of $\Bbb Z\oplus\Bbb Z$. This forces the top left arrow $\tilde{H}_2(X)\to\Bbb Z\oplus \Bbb Z $ to be an isomorphism, and $\gamma=0$. The top sequence then degenerates to a short exact sequence
$$0\to\Bbb Z \to\tilde{H}_1(X)\to \Bbb Z\oplus \Bbb Z \to 0$$
Consequently, $$\tilde{H}_1(X)\simeq \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$$
Your $H_3$ calculation is correct because $H_2(A\cap B)=0$. The $H_2$ calculation is a little fishy. Here is the relevant fragment of the MV sequence:
$$0\to H_2(X)\to H_1(A\cap B) \to H_1(A)\oplus H_1(B),$$
which turns into
$$0\to H_2(X) \to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2.$$
There is no reason to conclude that $H_2(X)=\mathbb Z_2$. Indeed, based on the topology, the map $\mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2$ should be given by $x\mapsto (x,x)$, which is injective. Hence $H_2(X)=0$.
But now you have enough to figure out $H_1(X)$. This is easiest if you use reduced homology. Then, similarly to what you've written you get
$$0\to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2\to H_1(X)\to 0.$$
Thus $H_1(X)\cong (\mathbb Z_2\oplus\mathbb Z_2)/\langle (x,x)\rangle\cong \mathbb Z_2$.
An alternative approach is to use cellular homology. This is basically $\mathbb{RP}^2$ with two $3$-cells attached. The cellular chain complex is
$$0\to \mathbb Z^2\overset{0}{\to} \mathbb Z\overset{\times 2}{\to}\mathbb Z\overset{0}{\to}\mathbb Z\to 0.$$
You can deduce these maps since you know what they are for $\mathbb{RP}^3$, and this is essentially $\mathbb{RP}^3$ with an additional $3$ cell attached in the same way. Taking the homology gives the same answer as Mayer-Vietoris.
Best Answer
The argument you've outlined for (b) works for (a) after a shift in perspective:
Suppose $L$ and $K$ are closed subspaces of $S^n$. Let $A=S^n- L$ and $B=S^n -K$. Then we have $A \cap B=S^n -(L \cup K)$, and the Mayer-Vietoris sequence reads $$H_{i+1}(A \cup B)\to H_i(S^n-(L\cup K)) \to H_i(S^n - L) \oplus H_i(S^n-K) \to H_i(A \cup B).$$ In part (b), you had $L=S^l$ and $K=S^k$ disjoint. This means $A \cup B=S^n$, so you get isomorphisms in the middle of the above sequence for all $i \leq n-2$. But in part (a), $L=S^l$ and $K=S^k$ intersect at a single point -- the wedge point. So we have $$A \cup B = (S^n - L) \cup (S^n - K) = S^n - (L \cap K)=S^n -\{pt\} \cong \mathbb{R}^n.$$
Leveraging the boring homology of $\mathbb{R}^n$ and the exactness above, we get isomorphisms $$H_i(S^n-(L \cup K)) \cong H_i(S^n - L) \oplus H_i(S^n - K)$$ for all $i\geq 0$. (The $i=0$ case can be argued using reduced homology, for example.) Now use the result from Hatcher again, just as in part (b).