= Solution
Because $S^n$ is simply connected for $n\geq2$, the homological <Serre spectral sequence> has constant coefficients and only two nonzero columns:
$$
E^2_{p,q}=H_p(S^n;H_q(F))
=\begin{cases}
H_q(F),&p=0,n,\\
0,&\text{otherwise}.
\end{cases}
$$
Its only possible nonzero differential is
$$
d_n:E^n_{n,i}\longrightarrow E^n_{0,i+n-1}.
$$
Using the orientation generator of $H_n(S^n)$ to identify both columns with $H_*(F)$ defines the <Wang homomorphism>
$$
\Delta:H_i(F)\longrightarrow H_{i+n-1}(F).
$$
The kernel and cokernel descriptions of the two surviving columns splice with the filtration of $H_*(E)$ to give the <Wang sequence over a sphere>
$$
\cdots\to H_j(F)\to H_j(E)\to H_{j-n}(F)
\xrightarrow{\Delta}H_{j-1}(F)\to H_{j-1}(E)\to\cdots.
$$
Let $F_m$ be the <homotopy fiber> of a degree-$m$ map $f_m:S^n\to S^n$. Apply this sequence to the fibration
$$
F_m\longrightarrow S^n\xrightarrow{f_m}S^n.
$$
At the bottom, the map between the two copies of $H_n(S^n)$ is multiplication by $m$. It follows that
$$
H_j(F_m;\mathbb Z)=
\begin{cases}
\mathbb Z,&j=0,\\
\mathbb Z/m,&j=k(n-1)\text{ for some }k\geq1,\\
0,&\text{otherwise}.
\end{cases}
$$
The first torsion group can also be seen from $\pi_{n-1}(F_m)\cong\mathbb Z/m$ and the <Hurewicz theorem>; the Wang map then propagates it periodically.
The <universal coefficient theorem for cohomology> gives
$$
H^j(F_m;\mathbb Z)=
\begin{cases}
\mathbb Z,&j=0,\\
\mathbb Z/m,&j=k(n-1)+1\text{ for some }k\geq1,\\
0,&\text{otherwise}.
\end{cases}
$$
Every product of two positive-degree classes is zero. For $n>2$, this follows immediately because the sum of two degrees of the form $k(n-1)+1$ is not of that form. For $n=2$, degree counting does not suffice, since $H^j(F_m)\cong\mathbb Z/m$ for every $j\geq2$. Use instead the multiplicative cohomological <Serre spectral sequence> for
$$
\Omega S^2\longrightarrow F_m\longrightarrow S^2.
$$
Its transgression in degree one is multiplication by $m$. Every positive integral cohomology class that survives lies in filtration two, while the product of two such classes lies in filtration four; the base has dimension two, so filtration four is zero. Thus the reduced cohomology is a square-zero ideal, and
$$
H^*(F_m;\mathbb Z)
=\mathbb Z\oplus\bigoplus_{k\geq1}(\mathbb Z/m)[k(n-1)+1]
$$
as a graded ring, with zero multiplication on the second summand.
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