As simple as in the title.. I would like to know also some mathematics about it!
It cannot. This is because energy and momentum are not both conserved if a free charged particle (say, an electron) emits a photon. It needs interaction with at least a second charged particle in order to do so (as in Bremsstrahlung). The mathematic involved is that of the energy of a photon $E=\hbar \omega$, energy of a particle $E^2 = m^2 c^4 + p^2 c^2$, momentum of a photon $p = \hbar \omega /c$ and simple trigonometry and basic algebra, very much as in the classical version of Compton scattering.
No. Photon does not change electron state ever.
Atom or molecule can have excited or no-excited state - depending how far electron is from nucleus.
According to nowadays view on the Space, it is not the emptiness. Space is heavily filled-in with waves (dark energy) and debris (like quarks, nutrino and other unknown yet particles - dark mass).
Nucleus makes a disturbance in the space and because of it creates spherical or more complex "wave boxes" (named shells or orbitals) around itself; for more than one boson in nucleus it can be cube-like-shape or some other wave combinations
Electron orbital in hydrogen is the gap between two spherical wave bumps of the space around nucleus; wave nature of electron makes orbital conductive and electrically atom does not radiate -- nucleus proton and electron shell are static to each other. Speed of electron is really does not matter and it makes random jerky moves inside the orbital. Keep in mind, orbital can have very different shapes. In molecule it is a tube between two nuclei.
Photon makes a disturbance in the atom adding electrical energy as a short pulse.
This energy pushes electron and proton apart; it is not always enough energy to push electron absolutely out of atom; it may simply push it from the lower orbital to the higher.
So the answer is: only atom, molecule or two particles with different electrical properties can emit/absorb photon. Photon is electromagnetic pulse and IS NOT a particle.