Let's start with the most elementary example: projective space $\mathbb P^n$. It's not hard to see that that the number of points on it is always $q^n + q^{n-1} + \dots + q + 1.$
Note that this is because $\mathbb P^n$ can be always decomposed into simpler pieces: $\mathbb A^n \cup \mathbb A^{n-1}\cup\dots\cup \mathbb A^0$. Interestingly, something similar applies to all $\mathbb F_q$-varieties. Specifically, the Lefschetz fixed points formula from topology applied to arithmetics gives the following statement for a variety $X/\mathbb F_q:$
There exist some algebraic numbers $\alpha_i$ with $|\alpha_i| = q^{n_i/2}$ for some $(n_i)$ such that the number of points $$\\# X(\mathbb F_{q^l}) = \sum_i (-1)^{n_i}\alpha^l_i\quad \text{for}\\ l > 0 .$$
Numbers $\alpha_i$ in fact come from geometry: they are eigenvalues of some operators acting on etale cohomology groups $H_{et}(X)$. In particular, the numbers $n_i$ can only occupy an interval between 0 and $\text{dim}\\, X$ and there are as many of them as the dimension of this group.
These groups can directly compared to the case of $\mathbb C$ whenever you construct your variety in a geometric way. To see how, consider the example of curves. Over $\mathbb C$ the cohomology have the form $\mathbb C \oplus \mathbb C^{2g} \oplus \mathbb C\ $ for some $g$ called genus; the same holds over $\mathbb F_q$:
- projective line $\mathbb P^1$ has genus 0, so it always has $n+1$ points
- elliptic curves $x^2 = y^3 + ay +b$ have genus 1, so they must have exactly $n + 1 + \alpha + \bar\alpha$ points for some $\alpha\in \mathbb C$ with $|\alpha| = \sqrt q.$ This is exactly the Hasse bound mentioned in another post.
These theorems, which provided an unexpected connecion between topology and arithmetics some half-century ago, were just the beginning of studying varieties over $\mathbb F_q$ using the geometric intuition that comes from the complex case.
You can read more at any decent introduction to arithmetic geometry or étale cohomology. There are also some questions here about motives which are a somewhat more abstract version of the above picture.
As a reply to Ben's comment above about reconstructing the genus if you know $X_n = \#X(F_{q^n})$:
You know with certainty that $1 + q^n - X_n = \sum \alpha_i^n\ $ for some algebraic numbers $\alpha_i, i = 1, 2, \dots $ having property $|\alpha_i| = \sqrt q.$
There cannot be two different solutions $(\alpha_i)$ and $(\beta_i)$ for a given sequence of $X_n$ because if $N$ is a number such that both $\alpha_i = \beta_i = 0$ for $i>N$ then both $\alpha$ and $\beta$ are uniquely determined from the first $N+1$ terms of the sequence.
So a given sequence uniquely determines the genus.
I don't know, however, if a constructive algorithm that guarantees to terminate and return genus for a sequence $X_n$ is possible. The first idea is to loop over natural numbers testing the conjecture that genus is less then $N$, but there seem to be some nuances.
In order to do geometry, you need to have some kind of global structure which has good local models (the "neighborhoods") and good gluing conditions. In algebraic geometry, the good local models are rings. If you want do geometry with a fibered category, you need gluing conditions (that is, you need your fibered category to be a stack), and you need local models, that is, you need your category to be locally, in some pre-topology, an affine scheme (this is not quite right, but I hope it gives a rough idea). The pre-topology must be such that if $X \to Y$ is a covering, the fact that $Y$ has certain "interesting" local properties implies that $X$ also has them. Étale coverings work very well, of course; smooth coverings also work, not quite as well.
So, you can't do geometry with the stack of coherent sheaves because this does not have good neighborhoods. See also my answer to Qcoh(-) algebraic stack? to see what can wrong.
As to why algebraic stacks are always assumed to be stacks in groupoids, there are several things I could say, but the honest answer is that I don't know the deep reason for this. I know that in practice it suffices, so there is no reason to give up the inversion map, which is quite useful. Just think of how much more you can say about group actions, than about actions of monoids.
Of course, this does not mean that in the future people will not feel the need to extend the theory of algebraic (or topological, or differentiable) stacks to the more general case.
[Edit]: So, why is a geometric stack a stack in groupoids? Well, the first reason is that the inversion map is very useful in proving results. Of course, if we needed to do without it, we would.
The second, more serious, reason, is that, in concrete examples, stacks with non-cartesian maps tend not to admit non-trivial map to spaces. For example, consider the stack $\mathcal M_{1,1}$ of elliptic curves. If we admitted all squares as morphisms, instead of only the cartesian ones, any map from $\mathcal M_{1,1}$ to a space would have to collapse an isogeny classes of curves to a point, and then one can see that it would map everything to a point. So, no moduli space.
As another example, take the stack of vector bundles on a projective variety $X$. There is a map between any two vector bundles, so no open substack could possibly admit a non-trivial map to a space.
Of course, if $F$ is a stack over a site $C$, there is substack $F^*$ with the same objects, whose arrows are the cartesian arrows in $F$; and if $X$ is an object of $C$, or a sheaf on $C$, any cartesian functor $X \to F$ would factor through $F^*$; so you could argue that a chart for $F$ would in fact come from a chart for $F^*$. In all the examples I know, $F^*$ is the right object to consider.
But, once again, none of these reasons is really compelling; for example, if monoid actions became important in geometry, I would bet that soon people would start working with geometric stacks that are not stacks in groupoids.
Best Answer
Here are two applications of stacks to number theory.
1) Section 3 of this paper, which solves the diophantine equation $x^2 + y^3 = z^7$, explains the connection between stacks and generalized Fermat equations.
2) This post explains how stacks fit into the proof of Deuring's formula for the number of supersingular elliptic curves over a finite field.