In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms.
A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a fermion. Bosons always have an integer spin; fermions always have a half-integer spin. By the spin-statistics theorem, the wave function of two bosons is invariant under their exchange but it is antisymmetric under the exchange of two identical fermions. These two rules are theorems for elementary particles and assuming this theorem, it's also trivial to prove these statements for composite particles.
In particular, for a neutral atom, the numbers of protons and electrons are equal so their total parity is even. That's why only neutrons matter. An isotope with an even number of neutrons is a boson (the whole atom: e.g. helium-4); an isotope with an odd number of neutrons is a fermion (the whole atom: e.g. helium-3).
That doesn't mean that composite bosons exhibit all the same physical phenomena like superfluidity that we may observe with some bosons.
The magnetic moment of a charged "elementary enough" particle scales like $1/m$ where $m$ is the mass of the particle. That's why the magnetic moment of the protons, neutrons, and nuclei are about 1,000-2,000 times smaller than the magnetic moments of the electrons. That's why the nuclear spins are largely negligible for the behavior of the atom in a magnetic field.
This is no contradiction because the whole atoms have a much larger magnetic moments than the nuclei separately – because of the neutrons: atoms are not "elementary enough" in this definition. Both the electrons' spins and their orbital angular momentum contribute to an atom's magnetic moment. Also, there exist a higher number of states because an atom is a typical example of the "addition of several angular momenta". The tensor product Hilbert space may be decomposed as a direct sum of Hilbert spaces with fixed values of the total angular momentum. The degeneracy and the magnetic moment of these components depend on the total angular momentum i.e. on the relative orientation of the nucleus-based and electron-related angular momenta.
In effect, the spectral lines of the whole atom exhibit the so-called hyperfine structure. Up to some approximation, the nuclear spin may be totally ignored. But when the considerations from the previous paragraph are properly account for, each spectral line is actually split to several nearby (3 or so orders of magnitude closer to each other) finer spectral lines, each of which corresponds to a different value of the total angular momentum of the whole atom (or, equivalently, a different value of $\vec J_{\rm electrons}\cdot \vec J_{\rm nucleus}$).
Quantisation does not imply discreteness. If a system has been quantised, we just mean we have taken the set of states, and replaced it by a vector space of states. In other words, one can add states in quantum mechanics, allowing a system to be in two states "at once".
Observable quantities become certain operators acting on this vector space of states.
As you can see, this doesn't have anything, on the face of it, to do with discretisation. It turns out, however, that a lot of the operators we are interested in have discrete eigenvalues, and this implies that the corresponding physical values are discrete. Position, however, has a continuous spectrum, as do many other quantum observables.
There are plenty of sources explaining exactly how one goes from classical sets of states and numerical observables to quantum states (vector spaces - Hilbert spaces in particular) and quantum observables (operators); I won't cover that. All I shall say is that quantisation is a big mathematical process replacing a load of classical things with quantum things, and this sometimes leads to certain physical quantities being discretised.
Best Answer
A fermion is any particle, elementary or composite, that obeys Fermi-Dirac (as opposed to Bose-Einstein) statistics relating to how identical particles behave when you swap two of them. Due to an important but complicated result, this is taken to amount to having half-integer spin.
A lepton is one type of elementary particle with spin 1/2. The only leptons are the three generations of electrons, the three generations of neutrinos, and the antiparticles of these. (So that's 6, 9, or 12, depending on how you count.) There are also elementary fermions (the quarks) that are not leptons, so this is a much narrower class of particles.