Smooth homotopy

The Whitney Approximation Theorem says that for smooth manifolds $X$ and $Y$ (with $\partial Y = \emptyset$), any continuous map $X \to Y$ is (continuously) homotopic to a smooth map $X \to Y$. Moreover, we can choose the homotopy to leave the map alone wherever it is already smooth. So in your case there is a smooth homotopy $F: M \times [0,1]\to N$ that is (continuously) homotopic to $f$ and satisfies $F_0=f_0$ and $F_1=f_1$.

Yes. I assume you mean $M$ and $N$ to be manifolds without boundary, though I expect the result is true when they are, too.

This follows from the fact that, given a manifold with (or without) boundary $f: (W, \partial W)$, a manifold without boundary $N$ and a map $(W, \partial W) \to N$ that's smooth on the boundary, $f$ is homotopic to a smooth map by a homotopy that doesn't modify the map on the boundary. This is known as the Whitney approximation theorem; you can find a proof in e.g. Lee's "Introduction to Smooth Manifolds". Simply take $W = M \times I$ and $f = f_t$; then by hypothesis $f|_{\partial W}$ is smooth.

Note that if we take $W = M$, we get that every continuous map is homotopic to a smooth one; so $[M,N]$ and $[M,N]_{\text{smooth}}$ are in bijection. So there's no interesting difference between "continuous" and "smooth" homotopy theory.