= Solution
Let $N=M_1\mathbin{\#}M_2$ and $A_i=M_i\setminus\operatorname{int}D_i$. Use open thickenings of the two pieces, whose intersection deformation retracts onto $S^{n-1}$. The <Mayer–Vietoris sequence>[Mayer-Vietoris sequence] then has terms from $H_k(A_1)\oplus H_k(A_2)$ and $H_k(S^{n-1})$.
For $n\geq2$, the pieces and intersection are <connected>. The map on $H_0$ from the intersection is $1\mapsto(1,-1)$ and is injective; it gives $H_0(N)=\mathbb Z$ and contributes no extra degree-one class. In intermediate degrees the <sphere> has no homology, except potentially at $n-1$. At that degree its map to $H_{n-1}(A_1)\oplus H_{n-1}(A_2)$ is zero because each oriented boundary <sphere> bounds the punctured piece's relative fundamental chain. Using $H_n(A_i)=0$, the top part of the sequence therefore gives $H_n(N)\cong\mathbb Z$ and $H_{n-1}(N)\cong H_{n-1}(A_1)\oplus H_{n-1}(A_2)$. The $H_0$ injection just noted handles the adjacent term when $n=2$.
Together with part (ii), the answer is
$$
\boxed{H_k(N;\mathbb Z)\cong\begin{cases}
\mathbb Z,&k=0\text{ or }k=n,\\
H_k(M_1;\mathbb Z)\oplus H_k(M_2;\mathbb Z),&0<k<n,\\
0,&k>n.
\end{cases}}
$$
For $n=1$, each <connected> <closed manifold> is a <circle> and each punctured piece is an interval. The intersection is $S^0$ with two components: its $H_0$ map to the two interval components has rank one and kernel $\mathbb Z$. Mayer-Vietoris gives $H_1(N)=H_0(N)=\mathbb Z$, so the same displayed formula holds, with no intermediate degrees. The usual <connected>-sum construction here uses positive dimension.
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