Why are the axial bonds of PF5 longer than those of the equatorial bonds? (Hybridization)
Older textbooks decided to use a d-orbital formalism for elements of the third and lower (in the periodic table) periods to explain hyperconjugation. Nowadays, most of these compounds are explained with multi-centre bonds.

I could go stealing pictures and explanations from ron’s great answer, but I’m not going to. Rather, I’m just going to quickly explain the gist and redirect you to that answer.

If you take away two of the fluorines, then you can imagine the $\ce{PF3}$ fragment to be sp²-hybridised with a remaining p-orbital containing two electrons. (This description is wrong, as can be seen from the pyramidial structure of $\ce{PCl3}$ and its analogues, but we’ll keep it as a working basis.) The p-orbital can now interact with two surrounding further fluorine atoms that approach it from above and below with an electron each in the way that is shown in the image in ron’s answer. The resulting three orbitals will thereby create a four-electron three-centre bond. This bond can also be described mesomerically, by assuming a $\ce{P-F}$ single bond, a positive charge on phosphorous and a negative charge on the other fluorine, with a second mesomeric structure as a mirror image of the first.

From this mesomeric consideration you can see that the bond order of the axial $\ce{P-F}$ bonds is not 1 but rather closer to 0.5. The greater the order the shorter the bond so the smaller the order the longer the bond — and there you have your result.