= Solution
Let $p:E\to S^r$ be the <Serre fibration>. For $r\geq1$, cover the base by $U=S^r\setminus\{s\}$ and a small open ball $V$ about $s$. Write $A=p^{-1}U$ and $B=p^{-1}V$. The contractible-base assumption gives $A\simeq F$ and identifies the pair $(B,A\cap B)$, up to fiberwise <homotopy>, with $(F\times V,F\times(V\setminus\{s\}))$.
<Excision> for the open cover $E=A\cup B$ gives $H_i(E,A)\cong H_i(B,A\cap B)$. The relative base pair has
$$
H_j(V,V\setminus\{s\};\mathbb Z)=\begin{cases}\mathbb Z,&j=r,\\0,&j\ne r,\end{cases}
$$
by the deformation retraction of the punctured ball onto $S^{r-1}$ and its <relative homology> sequence. The relative <Künneth theorem> therefore gives
$$
H_i(E,A;\mathbb Z)\cong H_{i-r}(F;\mathbb Z).
$$
There is no Tor term because the only nonzero base group is free. The <long exact sequence of a pair>, replacing $A$ by its homotopy-equivalent fiber, now becomes
$$
\boxed{\cdots\longrightarrow H_i(F;\mathbb Z)\longrightarrow H_i(E;\mathbb Z)
\longrightarrow H_{i-r}(F;\mathbb Z)\longrightarrow H_{i-1}(F;\mathbb Z)\longrightarrow\cdots.}
$$
The last arrow is the relative boundary transported through these identifications; it can encode twisting, so it has not been set to zero. This proves the <relative homology of a fibration over a sphere> sequence. In the dimension-zero case, the total space is the disjoint union of the two fibers and the same formula is the split sequence for that union.
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