On bonus question 3, for $f_k(p)=kp^2+1$, I have found several values for $k$ which have a 2-cycle.  I can't show that all $p$ will fall into these cycles or that these are the only cycles.  Working backwards from these cycles, we can find which numbers do fall into them.

For $k=1$ we have a cycle $(89,233)$;
$k=5$ has cycle $(43,67$); $k=11$ has cycle $(23,97)$; $k=23$ has cycle $(3,13)$;  $k=41$ has cycle $(67,409)$; $k=71$ has cycle $(5,37)$; $k=85$ has cycle $(383,769)$; $k=113$ has cycle $(509,1021)$; $k=143$ has cycle $(7,73)$; $k=215$ has cycle $(257, 467)$; $k=311$ has cycle $(67,277)$;  $k=359$ has cycle $(11,181)$; $k=863$ has cycle $(17,433)$; $k=951$ has cycle $(67,73)$; $k=1238$ has cycle $(13,41$); $k=1427$ has cycle $(167,293)$; $k=1808$ has cycle $(107,271)$; $k=1811$ has cycle $(17,61)$; $2089$ has cycle $(31,241)$; $k=2501$ has cycle $(167,251)$; $k=4199$ has cycle $(59,101)$; $k=5903$ has cycle $(61,151)$; $k=6119$ has cycle $(61,179)$; $k=6269$ has cycle $(29,67)$; $k=7679$ has cycle $(109,151)$; $k=9407$ has cycle $(7,97)$.

I found one 3-cycle at $k=7$ which goes $(79,127,1283)$.

As well, for $g_k(p)=kp^2-1$, aside from perfect square $k's$, I found $k=11$ has cycle $(239,3307)$; $k=19$ has cycle $(157,233)$; $k=23$ has cycle $(929,2711)$; $k=61$ has cycle $(127,503)$; $k=103$ has cycle $(857,1283)$; $k=1639$ has cycle $(127,197)$.

$k=123$ has a 3-cycle $(443,449,823)$; $k=409$ also has a 3-cycle $(71,317,137)$.

At first glance I notice how many of the $k's$ are prime, but not all.  Now, why so many 2-cycles?  I think it is because of my computational limitations since the numbers get large quickly.  I found these using unsigned long integer data type; I predict arbitrary precision calculations would uncover larger cycles.

For $k=23$, we have $23*3^2+1=2^4*13$, and $23*13^2+1=2^4*3^5$, and you can see that these large numbers must be highly composite to have such low greatest prime factors. Another pleasing example is $k=143$; $143*73^2+1=2^6*3^5*7^2$.  That explains why these are so rare.