Why is triplet carbene sp2-hybridised? Triplet carbene has a carbon with 6 electrons, of which 2 are unpaired. I would expect the carbon to be $\ce{sp}$ or $\ce{sp^3}$ hybridised and the two singly occupied orbitals to be degenerate. However, Clayden et al. (p. 1011) say that, since the triplet carbenes are almost always bent, they are best considered to be $\ce{sp^2}$ hybridised, with the unpaired electrons occupying a $\ce{p}$ and an $\ce{sp^2}$ hybrids . 

The very basic explanation I see why the two SOMOs are not degenerate is simply that $5E_{\ce{sp^2}} + E_{\ce{p}} + 2E_{\ce{sp^2},\ \text{pairing}}$ is less than $4E_{\ce{sp}} + 2E_{\ce{p}} + 2E_{\ce{sp},\ \text{pairing}}$ or $6E_{\ce{sp^3}} + 2E_{\ce{sp^3},\ \text{pairing}}$. However, I wonder if there's a better explanation.
That's actually a good question. I was lucky enough that Dietrich Gudat, who wrote a long chapter on this type of chemistry in the Riedel - Moderne Anorganische Chemie was my professor in inorganic molecular chemistry and we discussed these things in detail back then but it's been a while. 

I think it's due to how you want your final structure to look like or VSEPR. You have bascially three ligands, two rests and the lone-pair on the carbon. And by having it sp² you get quite some distance with the 120° angles. This splits your possible energy states in one which is called a1, that is basically the sp² and an b1 (the p-orbital). As the a1 has more s-character it is lower in energy and thus you can, depending on how big the gap is, either get a triplett or a singulett carbene.

This will change however if you move to the isolobal and isoelectronic nitrenes. Here we have no sp² but an sp which is automatically filled with two electrons. The other two electrons occupy the degenerate sets of px and py and thus you always end up with a triplett.