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Deriving photon energy equation using $E=m{c}^2$ and de Broglie wavelength
So I was learning de Broglie wavelength today in my physics class, and I started playing around with it. I wondered if it was possible to calculate the energy of a light wave given its wavelength and speed. After rearranging a bit, I plugged it into $E=m{c}^2$, and realized I had found the equation for the energy of a photon that I learned at the beginning of my quantum mechanics unit, $E=hf$
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
$p=\frac{h}{\mathrm{\lambda <!--\; \lambda \; -->}}$
$E=m{c}^2$
$E=\frac{p}{c}\cdot <!--\; \cdot \; -->c^2$
$E=\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}=hf$
I have a few questions about this. Firstly, I do not understand how it can make sense to do $\frac{p}{c}$ in this context, because, as I understand it, light has no mass. How can I come to $E=hf$ using the mass of a massless object?
Secondly, as I was doing those steps, I thought I would be calculating the energy of the entire light ray. I now realize I was finding the energy of a single photon. In hindsight, this makes sense, because the energy of the entire light ray must depend on some length value, correct?
This led me to two other questions: do light rays have some finite length? how do I calculate the energy of a light ray, not just a single photon?

De broglie equation
What is the de Broglie wavelength? Also, does the $\mathrm{\lambda <!--\; \lambda \; -->}$ sign in the de Broglie equation stand for the normal wavelength or the de Broglie wavelength? If $\mathrm{\lambda <!--\; \lambda \; -->}$ is the normal wavelength of a photon or particle, is $\mathrm{\lambda <!--\; \lambda \; -->}\propto <!--\; \propto \; -->\frac{1}{m}$ 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.)

De Broglie's Matter wave equation dividing by zero
I was just thinking about De Broglie's matter wave equation: $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$ where $p$ is the momentum of the object. But what if the object is at rest? Won't we be dividing by zero? What if we take the limit as momentum tends to zero, won't we start to get noticeable waves? Can someone please explain to me where I went wrong?