I am unsure on the parts $ii, iii, iv$. For $ii$, I am pretty confident that I pick the $4$ vertices in $nC4$ ways, order them in $4!$ ways, and then divide by $2$ as I have double counted due to (a,b,c,d) giving the same path as (d,c,b,a). So in all we have $\frac{4!nC4}{2} = 12\cdot nC4$.
I have no idea for $iii$ and $iv$ unfortunately because for these types of problems I only know how to count the number of paths or cycles in $K_n$. I will try though:
$(iii)$ Pick the $5$ vertices in $nC5$ ways.
$(iv)$ Pick the $5$ vertices in $nC5$ ways.
Formally, what you would need is the number of automorphisms of the labeled versions of each of these graphs, sometimes denoted by $\mathrm{aut}(G)$, e.g. in Random Graphs by Janson, Luczak and Rucinski. Let $v(G)$ denote the number of vertices in a (labeled) graph $G$. Then the number of copies of $G$ in $K_n$ is $\binom{n}{v(G)}\frac{v(G)!}{\mathrm{aut}(G)}$ because we choose the $v(G)$ vertices $\mathcal V\subseteq[n]=\{1,\dots,n\}$ and then one of the $v(G)!$ bijections $[v(G)]\rightarrow\mathcal V$. Finally, we notice that each of the resulting graphs on the vertex set $\mathcal V$ is attained under $\mathrm{aut}(G)$ of the $v(G)!$ bijections, by definition, hence we divide by $\mathrm{aut}(G)$.
So, the remaining challenge is to determine $\mathrm{aut}(G)$ for $P_3$, $S$ and $H$. For $P_3$, we label the vertices increasing from left to right and let $e_1=\{1,2\}$, $e_2=\{2,3\}$, $e_3=\{3,4\}$ be the edges. Since we are considering graph automorphisms, that is, bijections $\alpha:[4]\rightarrow[4]$ such that $\alpha(e_1)=\{\alpha(1),\alpha(2)\}\in\mathcal E=\{e_i:i\in[3]\}$ and the same for the other $e_i$, the degrees have to match. So, we have $\alpha(1)\in\{1,4\}$. For $\alpha(1)=1$ we necessarily have $\alpha(2)=2$ since $\alpha(e_1)=\{1,\alpha(2)\}\in\mathcal E$. Proceeding analogously, we notice that $\alpha=\mathrm{id}$ is the identity. For $\alpha(1)=4$ we observe analogously that $\alpha(v)=5-v$. Thus, we have $\mathrm{aut}(P_3)=2$.
For $S$, the degree condition above gives that the center of the star has to be mapped onto itself. The remaining vertices are then interchangeable, since no matter how we map them, they will again be connected to the center and only to the center. This gives $\mathrm{aut}(S)=4!$.
For $H$ we have only one degree $1$ vertex, say this is the vertex $5$. Let the unique neighbor of $5$ be $4$. Now, there exists exactly one vertex which is not adjacent to $4$, let this be $1$. Clearly, the remaining to vertices are interchangeable, since any way they will be exactly adjacent to $1$, $4$ and to each other. This shows that $\mathrm{aut}(H)=2$.