2 edited body; added 281 characters in body

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$which is an ample generator, and so we do have a canonical identification. This is trickier for other spaces, of course. 1 As far as when do spin structures exist, a manifold is spin if and only if the 2nd Stiefel Whitney class$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,$\mathbb{P}^n=Gr(2,n+1)$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.