I've only been thinking about (a) until now: Take a two sets of permutations, $A$ with $a_{n,k}$ elements s.t. each two have at least $k$ components in common and $B$ with $b_{n,k-1}$ elements s.t. each two have at most $k-1$ components in common. It's easy to show that $(a_i,b_i)\in A\times B, (a_1,b_1)\ne (a_2,b_2)\Rightarrow a_1\cdot b_1\ne a_2\cdot b_2$, so to each pair in $A\times B$ we can associate a different permutation in $S_n$. The number of pairs in $A\times B$ is $a_{n,k}\cdot b_{n,k-1}$, and $|S_n=n!$, and this proves what we want (in other words, we can define an injection $A\times B\to S_n$).

Part a) solves part b) we can choose (p-2)! permutations obviously., T0 prove that $a_p,2 \leq (p-2)!$ we find p(p-1) permutations that intersect in at most one entry . We group them into p- groups by their first entry. The entryies 2..p are cyclic permutations Consider permutations $/pi_i =1, 1+i,1+2i etc.$ (p is prime is crucial) We choose Them cyclically s.t. the first entries are different then do the procedure E.G. p=5 group 1: 1 2 3 4 5 1 5 2 3 4 (next cyclically) and so on We must only construct the p-th permutation, but this can be done by choosing the grouping in a certain way (for example we can let first element of $/pi_i$ be j not i) This helps to costruct a valid permutation for p-th group