= Solution
Write a point of the larger <sphere> as $(u,v)\in\mathbb R^{m+1}\times\mathbb R^{n-m}$. In the complement, $v\ne0$. There is a <homeomorphism>
$$
X\longrightarrow B^{m+1}_{\mathrm{open}}\times S^{n-m-1},
\qquad (u,v)\longmapsto\left(u,\frac{v}{\lVert v\rVert}\right),
$$
whose inverse sends $(u,w)$ to $(u,\sqrt{1-\lVert u\rVert^2}\,w)$. In particular the complement has a <deformation retraction> onto its transverse <sphere>. Explicitly,
$$
R_t(u,v)=\frac{((1-t)u,v)}{\sqrt{(1-t)^2\lVert u\rVert^2+\lVert v\rVert^2}},
\qquad 0\le t\le1.
$$
The denominator stays positive, the homotopy stays in the complement, and points with $u=0$ remain fixed. This establishes the <homology of an equatorial sphere complement>. By <homotopy invariance of homology>, its <homology> is therefore the <homology of a sphere> of dimension $d=n-m-1$:
$$
\boxed{H_j(X;\mathbb Z)=
\begin{cases}
\mathbb Z,&d\ge1\text{ and }j=0,d,\\
\mathbb Z^2,&d=0\text{ and }j=0,\\
0,&\text{otherwise}.
\end{cases}}
$$
In the codimension-one case, the complement has two contractible components; this is why degree-zero <homology> has rank two. Equivalently, in every case its only nonzero <reduced homology> is $\widetilde H_d(X;\mathbb Z)=\mathbb Z$.
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