Why does NF3 have a smaller bond angle than NH3?
This can be argued on the basis of Bent's rule; concisely stated


  Atomic s character concentrates in orbitals directed toward electropositive substituents


What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom.

Since s orbitals are lower in energy than p orbitals, they are better at stabilising electrons. Which is why molecules like to use them (s orbitals) in bonds where there is "more" electron density to stabilise.

Thus, Bent's rule can just as easily be formulated as follows:  p character concentrates in orbitals directed towards electronegative substituents; (the central atom doesn't have to waste its low energy s orbitals, the orbitals on the electronegative atom, can (possibly) take care of it.

What has all of this got to do with bond angles? Well, more s character leads to larger bond angles. You can think of how s orbitals are spherical in shape, and a large s character would lead to more "spherical" hybrid orbitals, and that would lead to larger bond angles.

Going back to your question, we are supposed to compare $\ce{NF3}$ and $\ce{NH3}$.

Fluorine is more electronegative that hydrogen, and the $\ce{N-F}$ bond would have greater p character than the $\ce{N-H}$ bond, (or $\ce{N-H}$ bond has greater s character, whichever you prefer). Thus $\ce{H-N-H}$ angle is greater in $\ce{NH3}$ than it is in $\ce{NF3}$



P.S I'm sure you can learn more about what I just tried to explain (rather poorly) by reading the Wikipedia article, and/or a textbook and/or using the search function on this site to find related questions/answers.