Like any quantum field you can approximate light as either a particle or a wave. However you need to be clear that both are just approximations to the true nature of light and like all approximations they work well in some circumstances and badly in others.
In this case the photon model is a poor way to describe the process of reflection. Reflection doesn't involve photons being absorbed then re-emitted. You're correct that the oscillating light wave of the light interacts with electrons in the reflector, and the resulting oscillating dipoles reradiate light. However, while this is easily described using a wave model it's hard to describe using photons. As a general rule the photon model works well when light is exchanging energy with something so for example it would be a good model if the light was ejecting photoelectrons from the mirror. When we're studying the propagation of light the wave model is a far better approximation.
So I don't think your question can usefully be answered as it's currently phrased. However we can say that the reflected light has approximately the same frequency as the incident light. A say approximately because in principle there are effects that could change the frequency. For example if the mirror is floating in zero-G then the momentum change when the light is reflected will cause the mirror to accelerate by a tiny amount and therefore red shift the reflected light by a tiny amount. In most circumstances we can ignore these small effects.
Another way to get atoms to emit light is to shine white light on an atom and the electrons would absorb the photon if the energy of that photon was equal to the energy difference between the energy levels. And the electron would jump to the next energy level and be absorbed. All other wavelengths do not have sufficient energy to allow an electron to jump to the next engery level so they will pass though the atom unchanged.
What is unclear to me is what you mean by being absorbed. As I say below, an electron cannot be absorbed, which is what I think you are implying above, but a photon, as the force carrier between electrons, can be absorbed and emitted.
I think there are duplicates for the other related questions in your post, so I will stick to the last two in this answer.
Why is it that the electron loses energy when it jumps to the next energy level
Let's take the common usage of the word jump as upwards. So in this above case, the electron gains energy. It loses energy when it falls back down to a lower level.
I have to admit that I don't like using words like jump and fall, because they are based on the Bohr model of the atom, which is not correct in almost every aspect.
So let me give you two pictures, one of the old model, which your question is based on, and one of the more modern picture.
The Bohr model (of 100 years ago)
The Orbital Distribution Density model
The electron will tend to lose energy if it can, by emitting a photon of the correct wavelength, that enables it to transition to a lower energy level, but if that lower level is already occupied to the maximum amount, then the electron is forced to stay at a higher level.
The difference between the pictures is the the Bohr model assumes a particle structure, whereas we now think in terms of the probability of finding an electron in a certain region, so we cannot be as definite as in the earlier model. Also, when the transition from one level to another occurs, it is not a smooth transfer like a car changing lanes, it is for a time a more chaotic operation, with the electron (or rather its' likelyhood of being found) bouncing around the place until it settles into a lower orbit.
In the first example the electrons moving with current gives energy to the electron in the atom. So the electron in the atom absorbs the moving electron? If so how is this possible because they are both negative?
There is no question of an electron absorbing another electron. Instead, by means of photon emission, momentum can be transferred between electrons, bearing in mind the conservation laws regarding energy and momentum.
An example of this is a Feynman Diagram:
Where the wavy line represents energy and momentum being transferred by means of a photon.
Best Answer
The physics behind these processes is captured in so-called cross-sections for photon-atom scattering. They can be formulated precisely given the Hamiltonian of the system and evaluated numerically. Important ones are elastic cross-sections, describing the situation where the photon is scattered from the atom but no energy is transferred, excitation cross-sections, where the photon is absorbed leaving the atom in an unstable excited state (which can decay to a state of lower energy emitting a second photon) and ionasation cross-sections, where the photon is absorbed and the atom has gained an energy above its ionisation energy causing it to decay into a positive ion and an electron. The interaction between the electromagnetic field (the photon) and the atom is the same in all cases but the values of the cross-sections depend on the photon energy. Note that although the photon energy may be sufficiently high to cause ionisation the probability for elastic scattering need not be exactly zero. Going beyond these simple processes, if the intensity of a photon beam is sufficiently high but the photon frequencies (i.e., their energy) are too low to cause ionisation then the atom can still be ionised by absorbing more photons (multi-photon ionisation). These processes are experimentally studied using intense laser beams. Here we encounter a cross-section for the process where initially we have an n-photon state (rather than one as above) and an atomic ground state.