Schubert calculus expressed in terms of the cotangent space of the Grassmannians

The tangent space to the Grassmanian corresponds to the following representation of $U(r)\times U(n-r)$, call it $\rho$: it is the $r\times (n-r)$ matrices, with $U(r)$ acting on the left and $U(n-r)$ acting on the right, so if we denote by $A$ the standard representation of $U(r)$ and by $B$ the standard representation of $U(n-r)$ we obtain $$ \rho = A\otimes B^*. $$ Now remember that the tangent space has Hodge decomposition into holomorphic and anti-holomorphic part, so the actual representation we are dealing with is $$ A\otimes B^* + A^*\otimes B. $$ So if we want to find invariant differential $k$-forms we need $U(r)\times U(n-r)$-invariants of $$ \Lambda^k(A\otimes B^* + A^*\otimes B) = \sum_{i=0}^k \Lambda^i(A\otimes B^*) \Lambda^{k-i}(A^*\otimes B). $$ Using the formula $$ \Lambda^n(A\otimes B) = \sum_{\lambda \vdash n} s_\lambda(A)\otimes s_{\lambda'}(B), $$ where $\lambda'$ stands for the conjugate partition, we obtain $$ \sum_{\lambda,\mu:|\lambda|+|\mu|=n} s_\lambda(A) s_{\lambda'}(B^*) s_\mu(A^*) s_{\mu'}(B). $$ Taking invariants we see that only terms with $\lambda'=\mu$ survive, and each one of them produces a one-dimensional space of invariants. Moreover, $s_\lambda(A)$ is non-trivial only if $\lambda$ has $\leq r$ rows, similarly $\mu$ has $\leq n-r$ rows. So we conclude that there is a one-dimensional space of invariants for each partition $\lambda$ in a $r\times (n-r)$ rectangle, and it is a one-dimensional space of differential forms of degree $2 |\lambda|$.

For your first question: yes. See Stoll, Invariant forms on Grassman manifolds, p. 15. I think your second question is answered in the same book.

Regarding your second question, I think the answer is in the famous Kostant "Lie Algebra Cohomology and the Generalized Borel-Weil Theorem" or rather its second part.