Degree of canonical bundle? Given a smooth complete intersection $X=D_{1} \cap D_{2} \cap \cdots \cap D_{k} \subset \mathbb{P}^{n}$ with ${\rm deg}\; D_i=d_i$, one can easily show that $\omega_{X} \simeq \mathcal{O}_{X}(\sum_{i=1}^{k} d_{i} -n-1)$, using induction on the number of hypersurfaces and the usual conormal sequence.
Here is the question.
Suppose $X \subset \mathbb{P}^{n}$ is a smooth projective variety of degree $d$, not necessarily a complete intersection. How to understand $\omega_{X}$ in terms of the embedding?
Is it even necessarily true that $\omega_{X}$ is restricted from a line bundle on $\mathbb{P}^{n}$?
Similarly, how to work out the cohomology of $\mathcal{O}_X$ and $\omega_X$? Does this only depend on the degree of $X$?
 A: Im not sure if this counts as a full answer, but it is a nice example which will hopefully shed light on some of your questions.
The canonical bundle $\omega_X$ of an Enriques surface $X$ satisfies $\omega_X \otimes \omega_X=\mathcal{O}_X$, but $\omega_X\neq \mathcal{O}_X$ in the Picard group. It follows that $\omega_X$ is not the restriction of any line bundle in $\mathbb{P}^n$, as these can't be non-zero torsion.
A: Smooth (or Gorenstein) subvarieties in $\mathbb P^n$ whose canonical bundle is a restriction from $\mathbb P^n$ are known as subcanonical, and are very special. A rational twisted cubic in $\mathbb P^3$ is not subcanonical, for obvious reasons of degree. 
It is most certainly not true that the cohomologies of $\mathcal O_X$ and $\omega_X$ only depend on the degree: for example, consider a twisted cubic as above and a plane cubic embedded in $\mathbb P^3$.
A: Take any curve at all, of any genus g, and any divisor of degree d > 2g. This embeds the curve into projective space with degree d, and a generic projection embeds it in P^3 also with any degree d > 2g. So d and n determine almost nothing about the curve.

On the positive side, interestingly, the nice counterexample given for the original question, a rational cubic in P^3, although not determined by its degree, is completely determined by its degree and the fact that (unlike the plane cubic) it spans P^3. (Rational normal curves are about the only examples I can think of, spanning but not a complete intersection, where d,n do determine all the invariants.) 
I guess you could give an inequality at least for the genus (i.e. h^1(O)) of curves in P^3, since a curve of degree d in P^3 projects to a plane curve of degree d-1, hence has genus bounded above by that of a general such plane curve. Indeed Castelnuovo has a famous such inequality.

