As pointed out by Francesco, part (1) is false in general; however, it is true when the first Betti number of $M$ is 0. Part (2) is correct. All this follows easily from Alexander duality, stating that if $d$ is the dimension of $m$, we have $\mathrm H_{d-1}(S^n) \simeq \mathrm H^{1}(M, M \smallsetminus S^n)$.

Of course, using this to show that  $S^2$ is not isomorphic to $D^3$ is a big overkill.