I have a few questions regarding the existence of a spin structure on Kaehlerian and hyperKaehlerian manifolds. I cannot seem to provide a reference for proofs or counterexamples, so references are more than welcome.
Q1: Does every hyperKaehler manifold admit a spin structure?
Q2: Does every Kaehler manifold admit a spin structure?
Q3: Are there any dimension constraints on the existence of a spin structure on such manifolds?
Remark: According to page 85 of Jost's Riemannian Geometry and Geometric Analysis (6th edition), every orientable Riemannian manifold in dimension 4 carries a spin$^c$ structure. Since complex manifolds are orientable, this offers a partial to answer to the above questions. Jost, however, provides no proof or reference.
A1: Yes, a hyperKähler manifold is spin. If we take our definition of a hyperKähler manifold to be a $4k$-manifold with holonomy group contained in the symplectic group $Sp(k)$ (see here) and we take our definition of a spin-manifold to be an $n$-manifold $M$ with a reduction of the frame bundle $Fr(TM)$ to $Spin(n)$.
We have the containments $Sp(k) \subset SU(2k) \subset Spin(4k)$ (see here and this question ). Therefore the reduction of the holonomy group of a $4k$-manifold to $Sp(k)$ is more than sufficient to reduce $Fr(TM)$ to $Spin(4k)$. Hence these are spin.
A2: No, a Kähler manifold need not be spin. Complex projective space is Kähler, but $\mathbb{C}P(2n)$ is not spin, as can be proven by computing that its second Stiefel-Whitney class is non-zero (see Milnor and Stasheff). I do not know if there are any nice results like "Kähler + ______ $\implies$ spin." It is worth pointing out that the $\mathbb{C}P(2n+1)$'s are spin (same reference).
I don't know anything about Q3.