Hybridization of heteroatom Why is the hybridization of oxygen in the seven membered ring pictured below $sp^3$, while the oxygen has a hybridization of $sp^2$ in the 5 membered ring? This also may be an error in the answer key as well.
It relates to the Hückel rule for aromatics. With the electron pair of the oxygen participating in the $\pi$ system of the oxazol (5-membered ring)  we get 6 $\pi$-$\ce{e-}$. Doing the same for the oxazepin would get 8 $\pi$-$\ce{e-}$.

Hückels rule states that planar, circular $\pi$ systems with $4n+2$ electrons are aromatic, resulting in a lower overall energy if the system is conjugated. (with $n$ being a positive integer or zero)
The other statement that Hückels rule does is that $\pi$ systems with $4n$ electrons are anti-aromatic, which avoid conjugation as it would increase the overall energy of the system.

Therefore it is favoured for your oxazol to conjugate, which results in a sp$^2$-hybridization for the oyxgen, while the oyxgen in the oxazepin avoids conjugation, resulting in a sp$^3$-hybridization.

The Hückel rule in turn can be quantized by a Hückel method computation or  graphically utilizing a Frost-Musulin-diagram.