How come you see what happens to the energy of a system when 2 free linear plane waves cancel out? Suppose we have 2 waves which perfectly cancel each other out, say, same frequency $f$ but with different phase (the second one has a difference of $\pm\pi$). Energies for both waves are the same $E=hf$. What does the sum of both $E'=2hf$ mean if there is no existent wave?
I answer below my personal impressions from what I have read here and here.
 A: If two waves cancel each other in particular points in space, the energy just gets redistributed to different points in space, like in a standing wave.
If two waves cancel each other everywhere, it means that they must originate at the same location, have the same amplitudes and opposite phases, which, practically, means that there are no waves to start with. 
A: Remember that from Planck's postulate, $E=hf$ stands for wave-particle matters, and that if 2 waves with same energy cancel out as you assume, the quantum mechanics interpretation as both of them free particles is a process of creation/annihilation of pairs after a collision, with a remnant of energy product of the kinetic energy before and after the collision (light photons and heat).
If you want to use $E=hf$ in a classical treatment, for 1D, then you can work out:
$$\Psi_1(x)=A\exp \left[i\left(kx-\omega t\right)\right]=A\exp \left[i\,2\pi\left(\frac{x}{\lambda}-\frac{Et}{h}\right)\right]$$
$$\Psi_2(x)=A\exp \left[i\left(kx-\omega t\pm\pi\right)\right]=A\exp(\pm i\pi) A\exp \left[i\left(kx-\omega t\right)\right]=-A\exp \left[i\left(kx-\omega t\right)\right]=-A\exp \left[i\,2\pi\left(\frac{x}{\lambda}-\frac{Et}{h}\right)\right]$$
$$\therefore \Psi_1+\Psi_2=0$$
Computing the total (kinetic) energy of both waves:
$$E=E_1+E_2=kA^2,$$
which does not say too much. But using a bit of what we learn from Modern Physics:
$$E_1+E_2=2hf=2\dfrac{h}{\lambda}v=2pv.$$
Notice that for $v=c$, one has the total energy of a photon using quantum mechanics (Planck-De Broglie's postulate $p=h/\lambda$) from a classical picture. Thus, for electromagnetic waves, you get that the total energy of the system would be that of 2 photons, which is emitted (absorbed) during the process.
A: Waves which perfectly cancel each other out are equivalent to no waves to start with. So the energy of the perfect cancellation will be zero. 
This does not violate conservation of energy. Note that if you start with two waves which don't cancel, you cannot bring them into perfect cancellation.
If you try to superimpose two coherent light sources with opposite phases, you will get a region where the beams cancel but there will be other regions where they interfere constructively. Overall, energy will be conserved compared to if the beams were not combined. The energies add as a sum, $E = E_1 + E_2$. This is not incompatible with energy being related to the square of $\mathbf{E}$ field amplitude, and amplitudes adding, because combining two sources will create regions of both constructive and destructive interference. 
$E=hf$ applies for a single photon. You cannot cancel two photons of opposite phases because photons do not have a phase. The classical waves that we are thinking about canceling are more closely associated with states involving many photons*. A single photon doesn't have an EM phase. Another way of putting this is the average value of electric field at any time for a state with 1 photon is 0 because it does not have a phase (the average of $|\mathbf E|^2$ can be nonzero). Adding a photon increases the energy by $hf$ but what does it mean to add a photon to a state? To really understand this we may have to use some quantum field theory. So perhaps adding the Planck relation to this discussion only over-complicates things.
*specifically they are superpositions of states with each state having a different number of photons.
A: If two waves exactly cancel then there is zero energy. It is a simple as that.
