 Solution

Let $N=M_1\#M_2$ and $A_i=M_i\setminus \mathrm{int}{D}_i$. Use open thickenings of the two pieces, whose intersection deformation retracts onto $S^{n-1}$. The Mayer-Vietoris sequence then has terms from $H_k(A_1)\oplus H_k(A_2)$ and $H_k(S^{n-1})$.

For $n\ge 2$, 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 isFor $n=1$, 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.