Identification of a compound C10H12O with 1H and 13C NMR data
Like you said, the 2 doublets at around $\pu{7 ppm}$ belong to a para-substituted benzene. The three single-proton multiplets around $\pu{5-6 ppm}$ belong probably to a single substituted alkene. A singlet and a doublet between $3$ and $\pu{4 ppm}$ belong to $\ce{CH3-{}}$ and $\ce{-CH2-{}}$ groups, respectively (Note that the signal that belongs to the $\ce{-CH2-{}}$ group is a doublet, indicating that it is adjacent to a carbon with only a single proton, so it is probably connected to the alkene).

The absorption maximum at $\pu{275 nm}$ gives us a hint that the disubstituted phenyl ring of our compound is not conjugated to the $\ce{C=C}$ bond.

Concerning the oxygen atom: it is definitely neither a carbonyl group (indicated by the absence of an intense peak at $\gt \pu{180 ppm}$), nor a hydroxy group. So it probably is an ether. 

So far, we have two possible candidates:


Allyl para-tolyl ether
1-allyl-4-methoxybenzene


The chemical shift of the $\ce{CH3}$-group is shifted downfield compared to the $\ce{CH3}$-shift in toluene (where it is around $\pu{2.3 ppm}$, while this compound has $\gt \pu{3.7ppm}$), and thus more likely belongs to $\ce{CH3-O-Ar}$ protons. So, our compound is 1-allyl-4-methoxybenzene.

However, the multiplets of the vinylic protons in the $\ce{^1H}$-NMR spectrum don't fit this compound well. The allyl group should give two dd signals for the terminal protons and a ddt for another vinylic proton. But the $J_2$ constant for terminal protons is very small ($\lt \pu{2 Hz}$) and can't be seen on some old NMRs, what would lead to the simple two doublets.

Why do we see dddd instead of ddt for $\ce{R-CH=CR2}$ proton? I'm not sure, but maybe this is because the $\ce{-CH2-{}}$ protons are diastereotopic. Then it makes sense.