Does photon behave as both a particle and a wave? Explain.

Calculating an energy of an electron with known De Broglie wavelength (why can't we calculate it similar than we do it for a photon)
Lets say we have an electron with known De Broglie wavelength $\lambda$. Can anyone justify or explain why we calculate its energy $E$ using 1st the De Broglie relation $\lambda =h/p$ to get momentum $p$ and 2nd using the invariant interval to calculate $E$
$\begin{array}{rl}{p}^2{c}^2& =E^2-{E_0}^2\\ E& =\sqrt{p^2{c}^2+{E_0}^2}\end{array}$
Why we are not alowed to do it like we do it for a photon:
$\begin{array}{r}E=h\nu =h\frac{c}{\lambda }\end{array}$
These equations return different results.

Proof of de Broglie wavelength for electron
According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$
The proof of this is given in my textbook as follows:
1.De Broglie first used Einstein's famous equation relating matter and energy,
$E=m{c}^2,$
where $E=$ energy, $m=$ mass, $c=$ speed of light.
2.Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,
$E=h\nu ,$
where $E=$ energy, $h=$ Plank's constant ($6.62607\times {10}^{-34}\mathrm{J}\mathrm{s}$), $\nu =$ frequency.
3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:
$m{c}^2=h\nu .$
$m{c}^2=h\nu .$
4.Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light:
$m{v}^2=h\nu .$
I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda =h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?

De broglie equation
What is the de Broglie wavelength? Also, does the $\lambda$ sign in the de Broglie equation stand for the normal wavelength or the de Broglie wavelength? If $\lambda$ is the normal wavelength of a photon or particle, is $\lambda \propto \frac{1}{m}$ true?
Can a wave to decrease its amplitude or energy with the increase of mass?
(My chemistry teacher told me that all matter moves in the structure of a wave and because of de Broglie's equation matter with less mass shows high amplitude and wavelength while matter with great mass shows less amplitude and wavelength. So as a result of that we see that objects like rubber balls, which have a great mass relative to the electron, move in a straight line because of its mass. However, I haven't found any relationship between wavelength and amplitude.)

Wien's displacement law in frequency domain
When I tried to derive the Wien's displacement law I used Planck's law for blackbody radiation:
$I_{\nu }=\frac{8\pi {\nu }^2}{c^3}\frac{h\nu }{e^{h\nu /k_{b}T}-1}$
Asking for maximum:
$\frac{d{I}_{\nu }}{d\nu }=0:0=\frac{\mathrm{\partial }}{\mathrm{\partial }\nu }(\frac{{\nu }^3}{e^{h\nu /k_{b}T}-1})=\frac{3{\nu }^2(e^{h\nu /k_{b}T}-1)-{\nu }^{3}h/k_{b}T\cdot e^{h\nu /k_{b}T}}{(e^{h\nu /k_{b}T}-1{)}^2}$
It follows that numerator has to be $0$ and looking for $\nu >0$
$3(e^{h\nu /k_{b}T}-1)-h\nu /k_{b}T\cdot e^{h\nu /k_{b}T}=0$
Solving for $\gamma =h\nu /k_{b}T$:
$3(e^{\gamma }-1)-\gamma e^{\gamma }=0\to \gamma =2.824$
Now I look at the wavelength domain:
$\lambda =c/\nu :\lambda =\frac{hc}{\gamma k_b}\frac{1}{T}$
but from Wien's law $\lambda T=b$ I expect that $hc/\gamma k_b$ is equal to $b$ which is not:
$\frac{hc}{\gamma k_b}=0.005099$, where $b=0.002897$
Why the derivation from frequency domain does not correspond the maximum in wavelength domain?
I tried to justify it with chain rule:
$\frac{dI}{d\lambda }=\frac{dI}{d\nu }\frac{d\nu }{d\lambda }=\frac{c}{{\nu }^2}\frac{dI}{d\nu }$
where I see that $c/{\nu }^2$ does not influence where $d{I}_{\lambda }/d\lambda$ is zero.

What is the most intense wavelength and frequency in the spectrum of a black body?
Planck's Law is commonly stated in two different ways:
$u_{\lambda }(\lambda ,T)=\frac{2h{c}^2}{{\lambda }^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}$
$u_{\nu }(\nu ,T)=\frac{2h{\nu }^3}{c^2}\frac{1}{e^{\frac{h\nu }{kT}}-1}$
We can find the maximum of those functions by differentiating those equations with respect to $\lambda$ and to $\nu$, respectively. We get two ways to write Wien's Displacement Law:
${\lambda }_{\text{peak}}T=2.898\cdot {10}^{-3}m\cdot K$
$\frac{{\nu }_{\text{peak}}}{T}=5.879\cdot {10}^{10}Hz\cdot K^{-1}$
We see that ${\lambda }_{\text{peak}}\ne \frac{c}{{\nu }_{\text{peak}}}$. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are ${\lambda }_{\text{peak}}$ and ${\nu }_{\text{peak}}$, how is ${\lambda }_{\text{peak}}\ne \frac{c}{{\nu }_{\text{peak}}}$?

Is there a significant error in using De Broglie's equation for an electron at really high speed?
I was wondering if using the De Broglie equation
$\lambda =\frac{h}{p}$
for object traveling at really high speeds would result in a significant error. For example if an object travelled at $0.02c$ would the error be negligible? How can I calculate the uncertainty in the result?

Wien's Displacement Law for real bodies
It is known that for perfect blackbodies,
$\lambda T=c$
where $\lambda =\text{peak wavelength}$
$T=\text{Absolute temperature}$
$c=\text{Wien's constant}$
But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses $\lambda T$ in terms of its emissitivity $ϵ$? I googled it, but found no relevant results.