= Solution
Identify $M=M_F^{(n-1,1)}$ with $F^n$, whose <basis vectors> record the singleton row. Put
$$
T=F(1,\ldots,1),\qquad U=\{(a_1,\ldots,a_n):\textstyle\sum_i a_i=0\}=S_F^{(n-1,1)}.
$$
If the <characteristic> does not divide $n$, then $M=T\oplus U$, and $U$ is simple. If it divides $n$, then $0<T<U<M$ is the <submodule> chain, with $T$ and $M/U$ trivial and $U/T$ simple of <dimension> $n-2$. In both cases there are no additional <submodules>: any vector with unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its <image> under $(i,j)$, and the <orbit> of such a difference spans $U$. A <submodule> containing no nonconstant vector is contained in $T$.
Now suppose $p\nmid n$. When $p\neq2$, part (b)'s valid vanishing $\operatorname{Hom}_{S_n}(M,W)=0$ implies $\operatorname{Hom}_{S_n}(U,W)=0$: extend a putative map using the <projection> $M\to U$.
When $p=2$, the assumption $p\nmid n$ makes $n$ odd. The corrected calculation in part (b) says every map $M\to W$ sends its singleton-row <basis vectors> to the same multiple of the globally <invariant> vector $v$. Hence it kills the augmentation <submodule> $U$. Extending a map $U\to W$ along the direct-sum <projection> again makes it zero. Thus, despite the characteristic-two error in the earlier claim, the requested conclusion remains valid:
$$
\boxed{p\nmid n\quad\Longrightarrow\quad\operatorname{Hom}_{FS_n}(S_F^{(n-1,1)},S_F^{(n-2,1,1)})=0.}
$$
The structural input is the <submodule lattice of the standard symmetric-group permutation module>.
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