The tangent bundle over a manifold is trivial iff the manifold is paralelizable

Question

Why is the tangent bundle over a manifold trivial if and only if the manifold is parallelizable?

What additional condition do we need to impose on a fiber bundle if we want it to be trivial exactly when it has $n$ nowhere vanishing smooth sections, where $n$ is a given natural number?

Answer 1

I'll answer the first part:

Because if it's parallelizable, you can find a global frame field of the manifold $M$> This frame field induces a bundle isomorphism to $M\times\mathbb{R}^n$ covering the identity on $M$. Conversely, if the tangent bundle is trivial, then a trivialization is a global frame field.