I'll start with the second question first.  Your question is to describe the tangent and cotangent bundles of projective spaces or a Grassmannian, and their exterior powers.  The Grassmannian $\text{Gr}(k,n)$ is also the Grassmannian $\text{Gr}(n-k,n)$, so you could say that it has two tautological bundles $A$ and $B$, of dimension $k$ and $n-k$.  Their direct sum is an $n$-dimensional trivial bundle.  My intuition from the geometry is that the tangent bundle is $\text{Hom}(A,B)$.  You can also write this as $A^* \otimes B$.

Now suppose that $k=1$ so that it is $\mathbb{C}P^n$.  Then $A = \mathcal{O}(-1)$ (if I have my signs right) and $B$ is the quotient of $(n+1)\mathcal{O}(0)$ by $A$.  So $A^* \otimes B$ is a quotient of $(n+1)\mathcal{O}(1)$ by $\mathcal{O}(0)$.  This is thus a resolution of the tangent bundle by direct sums of line bundles.  The cotangent bundle has a dual resolution, and you can explode a resolution like this to obtain a resolution of the exterior powers.

This realization of the tangent bundle of $\mathbb{C}P^n$ lets you compute its Chern classes.  I think that the values of the Chern classes tell you that it isn't more directly a direct sum of line bundles.  When $n=2$, Maple tells me that the Chern classes of the line bundles would be irrational.

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I imagine that this is discussed better in some more recent textbook, but Google finds [an old paper][1] by Hangan with the same statement that the Grassmannian tangent bundle is a tensor product.  The paper considers the geometric implications of having some tensor factorization of the tangent bundle of a manifold in general.

I would guess that you can't describe the tangent bundle of a general Grassmannian in terms of line bundles.  The line bundles yield a certain subgroup of the semigroup of possible total Chern classes.  I would think that this subgroup misses the Chern classes of the relevant bundles by a mile.  (But I don't know a rigorous argument or a reference.)

  [1]: http://www.springerlink.com/content/jt0761735m360h15/