As far as when do spin structures exist, a manifold is spin if and only if the 2nd Stiefel Whitney class $w_2(X)=0$, which is the same as $c_1(X)\mod 2=0$. So we must calculate $c_1(T_X)$.
Let $R$ be the universal subbundle and $Q$ the universal quotient bundle. Then we have $0\to R\to \mathbb{C}^n\to Q\to 0$ for $Gr(k,n)$, and the total Chern classes satisfy $c(R)c(Q)=1$ Thus, $c_1(R)+c_1(Q)=0$, and we can show that $c_1(R)=-1$ and thus $c_1(Q)=1$. But the tangent bundle is $\hom(R,Q)=R^*\otimes Q$, which means that $c_1(T)=n$, and completely ignores $k$.
So in particular, $\mathbb{P}^n=Gr(2,n+1)$ \mathbb{P}^n=Gr(1,n+1)$ has Chern class $n+1$, and so will be spin if and only if $n$ is odd.
I don't know the answers to 2 and 3.
Note: as Dave pointed out in the comment, I've identified $H^2$ with the integers for Grassmannians because there is a unique Schubert class $\sigma_1$ which is an ample generator, and so we do have a canonical identification. This is trickier for other spaces, of course.
