Hall subgroups of $\mathrm{PSL}$ The following is an exercise in Peter Cameron's notes on classical groups.

Exercise 2.10 (a) Show that $\mathrm{PSL}(2,5)$ fails to have a Hall subgroup of some admissible order.
(b) Show that $\mathrm{PSL}(2,7)$ has non-conjugate Hall subgroups of the same order.
(c) Show that $\mathrm{PSL}(2,11)$ has non-isomorphic Hall subgroups of the same order.
(d) Show that each of these groups is the smallest with the stated property.

I know how to solve these problems.
My questions:
(1) First consider $\mathrm{PSL}(2,p)$ where $p>3$ is a prime number. If $\mathrm{PSL}(2,p)$ satisfies each of the above conditions, what can I say about $p$?
(2) What can I say about $p$ if $\mathrm{PSL}(2,p)$ meets ALL the conditions (a)-(c)? Does such $p$ exist? (This is just (1), my bad. But how can I find $p$ such that (b) holds but (c) does not?)
(3) For a more general case, what if $p$ my first question is replaced by $q$, where $q>3$ is a prime power?
(4) How can we generalize (d) to $\mathrm{PSL}(n,p)$ or $\mathrm{PSL}(n,q)$ for some fixed $n$?
(5) The answer shows that (2) without (3) happens infinitely many times if we restrict our attention to a specific set $\pi$ of primes.  Is it still the case for the problem as originally framed (i.e., there are no non-isomorphic $\rho$-Hall subgroups for any set of primes $\rho$).
Any idea is a help. Thank you in advance!
 A: There are infinitely many prime powers $q$ for which (b) but not (c) can happen for $G := {\rm PSL}(2,q)$. This just depends on conditions of congruences of $|G|$ modulo powers of $2$, $3$, and $5$.
There are two mutually exclusive ways in which this can happen.
${\mathbf 1}$. If $8$ divides $|G|$, then $G$ has two conjugacy classes of subgroups isomorphic to $S_4$, but for some $q$ there are also subgroups isomorphic to the dihedral group $D_{24}$ of order $24$.
To get (b) but not (c), we need $S_4$ to be Hall subgroup of $G$. For this, we need $3$ but not $9$ to divide either $q-1$ or $q+1$, and similarly we need $8$ but not $16$ to divide $q-1$ or $q+1$. To avoid the $D_{24}$, we need either (a) $3|q-1$ and $8|q+1$; or (b) $3|q+1$ and $8|q-1$. 
This happens, for example, when $q = 7,23,41,103,137,151,281,\ldots$.
${\mathbf 2}$. If $60$ divides $|G|$ then $G$ has two conjugacy classes of subgroups isomorphic to $A_5$. To get (b) but not (c), we need these to be Hall subgroups of $G$ and also for there to be no subgroup $D_{60}$. 
I haven't written down the precise conditions for this in terms of congruences, bu it happens for example when $q = 11,29,131,139,211,229,331,\ldots$.
