Why is the first ionization enthalpy of molecular O2 less than atomic O?
A cousin of mine asked me to share her answer here:

In the molecular orbital diagrams shown in textbooks, the highest occupied molecular orbitals of $\ce{O2}$ and of $\ce{F2}$ are shown as antibonding. If the molecules and their cations had the same bond length, and the electrons in a species didn't &quot;talk to each other&quot;, and the orbital energies of an isolated atom were not influenced by the approach of another atom, you could predict that the ionization energy of the molecular species is lower than that of the atomic species. The argument would be that these HOMOs have higher energy than the corresponding atomic orbitals (see comments to question).
As these assumptions are not quite true, it is more complicated, but luckily the quick and dirty analysis yields the correct answer.
According to the MO diagrams, the bond order of $\ce{O2}$ is 2 and that of the cation $\ce{O2+}$ is 2.5 (after all, we are removing an antibonding electron). Indeed, the bond length of the cation is shown as smaller in this source. I did not find a bond length for the $\ce{F2+}$ species. This shows that you have to at least adjust the MO diagram for the change in bond length.
To get a rationale for the ionization energies in a quantitative manner, you could calculate the energies of the four relevant species ($\ce{O, O+, O2, O2+}$) and look at the differences.