Why/How alternating currents through coaxial cable do not interfere with each other?
I learnt from high school that all waves interfere with each other, and that is the principle of noise cancelling used in headphones.


... All waves on a potential field (say, the electric field in a coax wave guide, or the air) do superimpose, but they do not necessarily destructively interfere with each other.

Instead, model your signals \$s_1\$, \$s_2\$ and so on to be 

$$s_n(t) = A_n(t)\cos(2\pi f_n t + \phi_n(t))$$

and \$A_n(t)\$ and \$\phi_n(t)\$ to carry the actual information. We call these the amplitude and phase of the signal, and they can change over time (not much info to get through if they couldn't!). \$f_n\$ is your carrier frequency, i.e. the 300, 500, whatever MHz in your question. In reality, you'll have to assume \$f_n\$ is so high that for a couple of oscillations of the cosine, the amplitude and phase stay pretty much constant.

Now, your over receive signal would be \$r(t)\$,

$$r(t) = \sum\limits_n s_n\text.$$

Then it's quite easy to show that you can re-extract the individual \$s_n\$ without any influence of the other \$s\$. You can, for example, multiply your \$r\$ with a cosine of the one carrier frequency you care about, i.e. with \$x(t) = \cos(2\pi f_n t)\$.

Get out your trigonometric identity table from high school! \$\cos(a)\cdot\cos(b) = \frac12 (\cos(a+b)+\cos(a-b))\$, and thus

\begin{align}
s_n(t)\cdot x(t) &amp;=  A_n(t)\cos(2\pi f_n t + \phi_n(t))\cdot \cos(2\pi f_n t) \\
&amp;= \frac{A_n(t)}2(\cos(2\pi f_n t + \phi_n(t) + 2\pi f_n t) + \cos(2\pi f_n t + \phi_n(t) - 2\pi f_n t))\\
&amp;= \frac{A_n(t)}2(\cos(4\pi f_n t + \phi_n(t)) + \cos(\phi_n(t)))\\
&amp;= \underbrace{\frac{A_n(t)}2\cos(4\pi f_n t + \phi_n(t))}_\text{at twice the original frequency!} 
+ \underbrace{\frac{A_n(t)}2\cos(\phi_n(t))}_\text{around 0 Hz}\\
\end{align}

We can then just throw a low pass filter at this and get tha \$A_n\cos(\phi_n(t))\$ part alone – which contains all the information we put in there! 

All the other frequencies \$f_m, m\ne n\$ don't end up around 0 Hz with this method (but at \$|f_n \pm f_m|\$); so, they'd be filtered away by the same low pass filter.

What we just did was prove that for cosine oscillations, a method exists that can eliminate the presence of all other oscillations. Cosines are hence orthogonal when projected onto cosines in this way.

This is a bit of a very short introduction into basic signal theory. If the math was too much for you, remember this:

Different carrier frequencies are independent. Just like your noise-cancelling headphone can't cancel a 100 Hz tone with a 312 Hz tone, a 50 MHz transmission can without any problem be separated from a 600 MHz transmission.

Example:

\begin{align}
f_1&amp;= 5\,\text{MHz}\\
f_2&amp;= 10\,\text{MHz}\\
f_3&amp;= 20\,\text{MHz}\\
f_4&amp;= 100\,\text{MHz}\\
f_5&amp;= 300\,\text{MHz}\\
f_6&amp;= 500\,\text{MHz}\\
f_7&amp;= 650\,\text{MHz}
\end{align}

We want the signal at 300 MHz, so we mix with that:

\begin{align}
r(t) &amp;= \sum\limits_{n=1}^7 s_n(t)\\
 &amp;= \sum\limits_{n=1}^7 A_n(t)\cos(2\pi f_n t + \phi_n(t))\\
r(t)\cdot \cos(2\pi f_5t) &amp;=
\text{smth. at } f_1+f_5=305 \text{MHz and at } |f_1-f_5|= 295 \text{ MHz,}\\
&amp;\phantom{=}\text{ at } f_2+f_5=310 \text{MHz and } |f_2-f_5|= 290 \text{ MHz,}\\
&amp;\phantom{=}\text{ at } f_3+f_5=320 \text{, and } |f_3-f_5|= 280 \text{ MHz,}\\
&amp;\phantom{=}\text{ at }f_4+f_5=400 \text{, and } |f_4-f_5|= 200 \text{ MHz,}\\
&amp;\color{blue}{\phantom{=}\text{ at }f_5+f_5=600 \text{, and } |f_5-f_5|= 0 \text{ MHz,}}\\
&amp;\phantom{=}\text{ at }f_6+f_5=800 \text{, and } |f_6-f_5|= 200 \text{ MHz,}\\
&amp;\phantom{=}\text{ at }f_7+f_5=950 \text{, and } |f_7-f_5|= 350 \text{ MHz.}\\
\end{align}

If you now low pass filter this result with say, a filter that removes anything above 80 MHz, nothing survives but what was mixed down to 0 Hz. And that's coming from the frequency you cared about!