Why is the HOMO of reaction 3 raised in energy, compared to the others, if the carbocation has partial sp2 character due to resonance? Wouldn't the donation into the p-orbital be stabilizing thus lowering that orbital's energy? It's not like electrons are going into an antibonding σ* and π* orbital.
The intramolecular mixing from the oxygen non-bonding ($n_{OH}$) orbital in the carbocation into the empty $p$-orbital IS stabilizing and lowers its energy! That's the oxocarbenium cation! But it does not lead to the effect you propose.
It is great that you have the same case but without the -OH group so we can compare. In the image below, I added the effect of the lone pair $n_{OH}$ orbital in the system. When the electrons in this orbital mix with the vacant $p_{C+}$ orbital, they get stabilized. This too stabilize the carbocation as a whole. The amount of stabilization of the bonding orbital ($n_{OH}+p_{C+}$) and destabilization of the anti-bonding ($n_{OH}-p_{C+}$) depends on the energy match between the initial orbitals $n_{OH}$ and $p_{C+}$. And these amounts are different; usually the destabilization is greater than the stabilization. These rules also apply when the mixing is intermolecular, like in the case of a nucleophile's HOMO and an electrophile's LUMO.
And if we check closely since we lowered the $n_{OH}$ of the carbocation mixing it with $p_{C+}$, the energy of the carbocation's LUMO ($p_{HOC+}=n_{OH}-p_{C+}$) got higher!, which widens the energy gap with the butene's HOMO compared with the case without the -OH substituent (by about the difference between the dotted blue line and the first green one), so the interaction is not as good now and the bonding orbital $\sigma_{c-c}$ doesn't get as stabilized. Resulting in being a little higher than the one in the first reaction of your original image.
P.S. where did you find that image? I would like to check out the source, looks interesting.
Two possible ways to approach this:
System stability is effectively determined by the stabilization of electrons. But if the orbital is empty, there is no direct stabilization. But if the system is stabilized in the occupied orbitals, then things must be destabilized in terms of the empty orbitals.
Alternatively, the third example involves an oxygen in the bond. Given the increased effective nuclear charge, you'd expect some additional stabilization in the bonding electrons, so again, that might point to additional destabilization of the antibonding orbitals.
