I find this question quite confusing myself, because it is over-simplified. There is a lot which is not explained. However, I think we must interpret the diagram very simply, because that is how it is presented.
I presume that -5.0eV corresponds to the "ground state" (lowest energy level) of the atom and the other levels correspond to the possible energy levels which can be reached from it, 0.0eV corresponding to an electron being freed (unbound) from the atom. The energy levels are negative because energy must be given to the atom to excite it. Absorbing the least energetic photon would excite an atom from the -5.0eV level to the -4.0eV level. I expect you can calculate the wavelength of such a photon (Part B).
Part A is much more complicated, because it involves several issues : partial absoption of energy, absorption followed by emission, and possibly also consecutive absorption/emission events. Part B involves absorption only. Part A also has several possible answers.
The difference between bombardment with electrons instead of photons is that the whole of the electron kinetic energy does not have to be absorbed. So an internal transition (if one occurs) can be less than the KE of the bombarding electron; the remaining KE can be retained by the bombarding electron, or taken up as KE of the whole atom. The atom can be excited to any level which is no more than 3.7eV above the ground level - ie to levels 1, 2 or 3 (counting the ground level as 0). The excited atom then loses energy by emitting one or more photons - ie transitions $3\to 2/1/0$, also $2 \to 1/0$ and $1\to 0$.
A complication is that an excited atom could gain energy from a second bombardment and be excited above level 3 (-2.5eV). Photons can then be emitted from transitions $5 (0.0eV) \to 4/3/2/1/0$ and $4 (-0.5eV) \to 3/2/1/0$. However, you are probably not expected to consider the possibility of a 2nd excitation.