It is false in general that "abelian intersect maximal implies normal": $A_5$ is maximal in $S_5$, the subgroup generated by $(1,2,3,4)$ is abelian, but the intersection of the two is nontrivial (contains $(1,3)(2,4)$) and not normal in $S_5$.
However, it is true that the intersection of a maximal subgroup and an abelian normal subgroup is normal.
Proposition. Let $G$ be a group, and $H$ a maximal subgroup of $G$. If $N$ is an abelian normal subgroup of $G$, then $N\cap H$ is normal in $G$.
Proof. If $N\subseteq H$, then $N\cap H = N\triangleleft G$ and we are done.
If $N$ is not contained in $H$, then maximality of $H$ and normality of $N$ imply that $HN=G$ (since $HN$ is a subgroup). Let $x\in H\cap N$ and $g\in G$. Then we can write $g = hn$ with $h\in H$ and $n\in N$. Then
$$gxg^{-1} = (hn)x(hn)^{-1} = h(nxn^{-1})h^{-1} = hxh^{-1}$$
with the last equality since $N$ is abelian. Now, $x,h\in H$, so $hxh^{-1}\in H$. And $x\in N$, so $hxh^{-1}\in N$. Thus, $hxh^{-1}\in H\cap N$, proving that $H\cap N$ is normal in $G$. QED
Added. More generally: note that $H\cap N$ is certainly normal in $H$. If $HN=G$, then you only need to show that $N$ normalizes $H\cap N$: then $(hn)x(hn)^{-1} = h(nxn^{-1})h^{-1}$, which will lie in $H\cap N$ if $nxn^{-1}\in H\cap N$. Therefore:
Proposition. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. Then $H\cap N\triangleleft HN$ if and only if $N\subseteq N_{HN}(N\cap H)$, where $N_{NH}(N\cap H)$ is the normalizer of $N\cap H$ in $NH$.
Proof. If $N\subseteq N_{NH}(N\cap H)$, then the argument proceeds as above. Conversely, if $N\cap H\triangleleft NH$, then $N\subseteq NH=N_{NH}(N\cap H)$. $\Box$
In particular, if $N$ is abelian then $N\subseteq C_{NH}(N\cap H)\subseteq N_{NH}(N\cap H)$; and if $H$ is maximal, then this gives the proposition above in the nontrivial case.
Now apply this to $H$ and the abelian normal subgroup $G^{(n-1)}$ to conclude that $H\cap G^{(n-1)}\triangleleft G$.
Now, you know that $G^{(n-1)}$ is normal in $G$ (because the derived terms are always normal in $G$), and has order $p$. Since a $p$-Sylow subgroup of $G$ has order $p$, then $G^{(n-1)}$ is a $p$-Sylow subgroup of $G$. Since it is normal in $G$, the $p$-Sylow subgroup of $G$ is normal in $G$. So $G$ is contained in the normalizer of the $p$-Sylow subgroup $G^{(n-1)}$ of $S_p$.
Two subgroups of order $7$ are either equal or intersect trivially. Indeed, their intersection is a subgroup of each of them, and they have exactly two subgroups.
In general, a group $G$ can very well have many $p$-Sylow subgroups intersecting non-trivially. For example, let $H$ be any group which does not have a normal $p$-Sylow, and consider the group $G=\mathbb Z_p\times H$.
Best Answer
For your first question: Let $\sigma \in S_p$. Then $\sigma$ can uniquely be written as a product of disjoint cycles. The concatenation of the lengths of these cycles is called the cycle-type of $\sigma$, for example, the cycle type of $(12)(34)$ in $S_5$ is $2$-$2$-$1$. Let $m_1$-..-$m_k$ be the cycle type of $\sigma$. Then $\text{order}(\sigma)=\text{lcm} (m_1,..,m_k)$. This can be seen by noticing disjoint cycles commute, so $\sigma^k=\text{id}$ if and only if $k$ is a multiple of $m_j$ for all $j$. Now suppose $\text{order}(\sigma)=p$, then $p=\text{lcm}(m_1,..,m_k)$ where $m_1$-..-$m_k$ is the cycle type of $\sigma$. Since $p$ is prime, $p$ is only a multiple of $1$ and $p$, so all $m_j$ are either $1$ or $p$. Obviously they cannot be all $1$ so one must be $p$, but then $k=1$ since $\sum_{i=1}^k m_i = p$. So the cycle type of $\sigma$ is $p$ and the result follows.
For your second question: suppose $H$ and $K$ are subgroups of $S_p$ of order $p$, $H \neq K$. Then $H \cap K$ is a subgroup of $H$, so its order is a divisor of $p$. If its order is $p$ then $H \cap K = H = K$, so its order must be $1$. But this implies $H \cap K$ is the trivial group.