As has been described in earlier answers, a circle can be divided by compass and straightedge into $n$ equal parts if and only if $n=2^a p_1p_2\cdots p_k,$ where $a$ is a non-negative integer and the $p_i$ are distinct Fermat primes. (We allow the possibility $k=0$.)
The Fermat primes are the primes of the form $2^{2^m}+1$. There are only five Fermat primes known, $3$, $5$, $17$, $257$, and $65537$. A large amount of work has been expended to find additional Fermat primes, so far without success. It has even been conjectured, with some heuristic justifications, that there are no Fermat primes beyond these five.
Thus we have an essentially complete answer for the "equal parts" part of your question.
What about unequal parts? One can give an answer, but it is not nearly as satisfactory. Let $A$ and $B$ be points on a circle, for definiteness the unit circle. Let $d(A,B)$ be the (shorter) distance between $A$ and $B$ along the circle. The ratio $d(A,B)/\pi$ tells us what "fraction" of the circumference of the circle is taken up by the shorter arc joining $A$ and $B$. Then we have the following result.
Given a real number $r$, with $0\le r \le 1/2$, there exist constructible points $A$ and $B$ such that $d(A,B)=r$ if and only if the number $r$ is Euclidean.
A real number $x$ is Euclidean if it can be obtained, starting with the number $1$, using a finite number of additions, subtractions, multiplications, divisions, and applications of the square root function.
There is a very detailed analysis of all these matters, with proofs, in these University of Utah notes. All the equal parts stuff is there, and much more. Inevitably, the details require a certain amount of algebra. However, the presentation does not use Galois Theory.
Commensurable parts: Two arcs on the unit circle are called commensurable if the ratio of their lengths is a rational number. We can ask for a characterization of the possible straightedge and compass divisions into parts that are all commensurable with the circumference of the circle. It turns out that we can divide a circle into commensurable parts if and only if the ratio of each part to the full circumference is of the shape $\frac{m}{2^a p_1p_2\cdots p_k}$ where $m$ is a positive integer, and, as before, the $p_i$ are distinct Fermat primes. The proof is straightforward, if we take for granted the characterization of those $n$ for which the regular $n$-gon is constructible.