The sign $\lambda$ in de Broglie's equation,
$$\lambda=\frac h p,$$
is indeed the de Broglie wavelength of the object involved. It is the only wavelength one can meaningfully give a material particle, and "normal wavelength" is meaningless in that context. de Broglie's relation is also true for a photon (though it amounts to a calculation of a photon's momentum), where the wavelength is the usual wavelength of light.
Though this is a flawed picture, you can see de Broglie's relation as describing a "wave of probability" associated with the material particle. This wave must have total "weight" equal to $1$, which means that the particle must be somewhere. (The relative "weights" of the wave in different volumes give the relative probabilities of us finding the particle there.) Particles with higher mass have smaller de Broglie wavelengths, so they can be localized better. If they are located in smaller volumes their amplitudes (specifically: their probability densities) must be bigger, to keep the total "weight" equal to $1$.
The amplitude of this "wave of probability" is of course not related to the energy - that has more to do with its momentum and therefore with its wavelength.
Yes the product $\nu \lambda$ makes sense as a velocity. Defining $E = \hslash \omega$ and $p=\hslash k$ (the Planck constant $h=2\pi \hslash$, where the $2\pi$ is injected into the $\hslash$, since physicists usually prefer to discuss the angular frequency $\omega=2\pi\nu$ and the wave vector $k=2\pi p$ rather than the frequency $\nu$ and the momentum $p$ for Fourier-transform notational convenience), you end up with
$$\nu \lambda = \frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$$
which is the standard definition for the phase velocity. It corresponds to the velocity of the wave component at frequency $\nu$ propagating over distance $\lambda$ per unit time.
This is always true, but for more complicated situation, a system is represented by a superposition of different waves propagating at different phase velocity. It results a wave-packet propagating, as an ensemble, at the group velocity
$$v_{g}=\frac{\partial \omega \left(k\right)}{\partial k}$$
where the dispersion relation of the wave-packet is noted $\omega \left(k\right)$.
Only the group velocity has some physical clear interpretations. For instance, the phase velocity can be larger than the speed of light, but the group velocity can never been larger than the speed of light $v_{g}\leq c$, at least in vacuum.
For a photon in free space for instance, $\omega \left(k\right)=c k$ and thus it's group velocity is $c$. For a free non-relativistic particle of mass $m$, $\omega = \hslash k^2/2m$, and $v_{g}=\hslash k/m$. etc...
For a human body, you have to count all the atoms constituting the body. The individual frequency of one atom interferes with the frequencies of all the others, resulting in an almost flat dispersion relation. The group velocity is then ridiculously small. A human body does not move due to quantum effect ! You can get a basic idea of the group velocity of a human body supposing you are a free particle of mass $100 \textrm{kg}$ and wavelength $1 \textrm{m}^{-1}$ (i.e. the order of magnitude of your size is about $1 \textrm{m}$) the $\hslash$ kills $v_{g} \sim \left(10^{-36}-10^{-34}\right) \textrm{m.s}^{-1}$ !
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Best Answer
1) Yes. The photon's matter wave is actually its electromagnetic wave.
2) Photon emission is not such a kind of process when you get some particles emitted in some interval of time, and you can assume some emission moment within that interval to every particular particle. No, it is a quantum process. The system's intermediate states are not "some particles are emitted, and some are not emitted yet", but "system has some intemediate probability to be in the initial state, and some probability to be in the final state". Every photon is the subject to that accumulating probability. So the most we can say for every photon is that it is emitted in the same interval of time.
The frequency, polarisaion, direction, spatial distribution and all such characteristics of each photon would be the same as those of the electromagnetic wave. (Some advanced details are omitted.)
3) The same way as you may interpret some particular waveform in different bases, like a function of time or a set of sinewave weights, you can interpret photons in different bases. The non-basic waveform is then understood as a quantum superposition of states that belong to the chosen basis.
4) The question is based on the wrong assumption, see the answer to the question 2.
P.S. You deleted your 4 questions, leaving just one, but I hope my answers would still help you.