There are local obstructions to existence of crepant resolutions of quotient singularities, for example none exist for the Gorenstein quotient singularity $\mathbf{C}^4 / \pm 1$, see mathoverflow.net/questions/66657/….

I found that the main Proposition is the result of Katz from "Small resolutions of Gorenstein threefold Katz" (mentioned by Hailong Dao in the comments), who also does the $cD_n$ case. Katz interprets cDV singularties and their resolutions in terms of simultaneous resolutions of surface singularities and somehow reduces to the universal case $xy = (z-a_1 w) \cdots (z-a_{n+1} w)$.

I think an isolated $cA_n$ singularity $xy + f(z,w)$ admits a crepant (= small) resolution if and only if $f(z,w)$ is a product of smooth curve germs. Existence is inductive blow up of ideals such as $(x,h(z,w))$ where $h$ is a factor of $f$. The converse requires some justifcation...

Regarding EDIT 3-3: the fact that a degree 60 monomial is not a product of degree 30 monomials is showing that $\mathcal{O}(a)$ does not give a projectively normal embedding, that is the polytope is not normal. However the fact that it's not very ample (which is also true) is a more subtle check, right?

Can one do induction on dimension, using the following: intersecting $X$ with a general hyperplane through a nonsingular point $P \in X$ is also nonsingular at $P$? The base of induction would be that a generically reduced plane curve has a line which is nowhere tangent.

Philip Engel: most likely PD won't hold in this case. E.g. for threefolds you have 4 rays in maximal cones, and I think $H_2(X)$ and $H_4(X)$ will typically have different dimensions. One way to approach this is to subdivide the cones to make an orbifold resolution $X' \to X$ and compare homology of $X$ and of $X'$.

Not sure from the question if you want integrally or rationally? An easy condition is that if for every cone in the fan the rays are linearly independent (I think such cones are called simplicial), then the toric variety has quotient singularities and Poincare duality holds rationally.

Welcome to Mathoverlow! Regarding the first sentence of the question, why are you saying that a general element of a linear system is irreducible? Clearly if your linear system contains just one hypersurface, which is reducible, then it's false? Bertini's theorem will tell us that they are generically irreducible provided the system is sufficiently moving, e.g. basepoint free...