Finite groups of Lie type include those obtained as rational points of a connected simple algebrraic group over a finite field $k = \mathbb{F}_q$ of characteristic $p$: these are split or quasi-split. There are also several families obtained less directly from algebraic groups: the Suzuki groups $^2\!B_2(q)$ and the Ree groups $^2\!G_2(q),\: ^2\!F_4(q)$, where $q = p^{2n+1}$ with $n \geq 1$ is an odd power of (respectively) $2,3,2$. [Caution: Sometimes $q$ is written here as $q^2$ to emphasize the similarity of group orders to those of untwisted groups.]
Counting the total number of conjugacy classes in such a group is a natural problem, relative to the determination of ordinary characters. Case-by-case study has been done for many of the families, especially the exceptional types; but it's unclear how much can be expressed uniformly. (Older results are surveyed in Chapter 8 of my 1995 AMS book on conjugacy classes.) It helps to assume the ambient algebraic group is simply connected, in which case Steinberg proved for split and quasi-split types that the number of semisimple classes (whose elements have order prime to $p$) is $p^r$ with $r$ the Lie rank. For Suzuki or Ree groups, $r$ is the rank of the BN-pair: 1, 1, 2.
For Suzuki groups the total number of classes is $q + 3$ (Deriziotis), while for Ree groups of type $G_2$ the number is $q+8$ (Ward). The latter groups came up recently here.
How many conjugacy classes do the Ree groups of type $F_4$ have?
This should be computable from known data, though not easily. Judging from all special cases I'm aware of, the answer should be given by a polynomial $q^2 + aq + b$, where $a, b \in \mathbb{Z}$ are independent of $q$. [Note however that for most other families of groups of Lie type, there are extra complications related to isogeny type, bad primes, and such.]
The underlying question, of course, is whether one can predict a priori what the polynomials will look like for all groups (coming from simply connected algebraic groups) starting with the highest degree term $q^r$.