Photons interact with matter if the matter offers quantum transitions that match, or nearly match, the photon's energy in the inertial frame of the matter. Ordinary matter such as wood, stone, etc. offers several groups of possible quantum transitions.
- Rotation of molecules (if they are free to rotate, i.e., not condensed matter)
- Vibration of molecules - bending, quivering actions
- Electronic excitations
- Nuclear excitations (there being various kinds, ignored here for simplicity)
Microwaves have such low energy they can't do much, though they might excite some types of vibrations on larger floppier molecules - however, any type of molecule that could be described as "floppy" probably isn't good for construction materials. Rotational modes aren't possible in a strong material made of crosslinked polymers or silicates. So microwaves mostly fly right through.
Near-infrared and visible light can kick electrons into higher molecular orbitals. Even if the energies aren't a match, just close, there is interaction, as Heisenberg lets them cheat temporarily. Also, having more energy, visible light photons can stir up a greater variety of vibrational modes. There's nothing in common wall materials to prevent that, and in fact, the interaction with photons is so strong that the material, if not super-thin (microns), will be opaque. Of course, glass is an exception.
Gamma rays are of such high frequency, electrons (or ions, or polarized ends of molecules) can't keep up due to inertia - so no interaction, or only a little. At the right frequencies, gamma photons can interact with nuclei, but for a randomly chosen source of gammas, its photons are unlikely to match closely enough with any of the available nuclear excitations, and can't really do much at the molecular level - therefore, the material is almost transparent.
All this is so oversimplified...
Background:
Einstein's photoelectric effect theory won him the Nobel prize, and it relates very closely to this. Although it's different from photoionisation, it relies on similar ideas.
His proposition: each atom will absorb the energy of one photon, and the energy of a photon is given by $$E=h\nu$$where $h$ is Plank's constant, a really, really tiny number ($6.62607004 × 10^{-34} m^2 kg / s$), and $\nu$ is frequency of the light. Higher intensity light, which is analogous to wave amplitude, contains more photons, but the energy of each photon is the same for a given frequency.
If I shine low-frequency high-intensity light on a surface, there's plenty of energy, but each atom, upon absorbing one photon, won't be able to lose an electron. However, if I shine high-frequency light, even if the intensity is low, each atom which absorbs a photon will be able to loose an electron, and we see ionization.
But when the intensity is high enough, even low frequencies will cause ionization.
This phenomenon, called multi-photon ionization, occurs when the atom absorbs more than one photon. It's usually pretty rare, because an atom frequently emits the other photons before it absorbs enough energy totally, but at very high enough intensities, it's appreciable.
Sound works differently in air: we generally don't say it's quantized in the same way, although if you examine it more minutely, you'll see that it can be quantized as phonons, which aren't evident in gases. But that's not relevant to your question, it's just something to keep in mind if you want to generalize a bit more by discussing sound in condensed matter.
Oddly enough, the parallel to sound and hearing loss is wrong! See this Biology SE question... high-frequency sounds are dangerous not particularly because they have more energy (which they do, see the equation for sound energy in a container), but because of the nature of the human ear and the alignment of hairs in it.
Best Answer
Very Low Frequency (VLF) and Extremely Low Frequency (ELF) radio waves are used for communication between submarines, because higher-frequency radiation doesn't propagate well through sea water.
I think the main reason they're not used much outside of the water is simply that higher frequencies are easier to use for a number of reasons. For one thing, the lower the frequency you're trying to use, the longer your antenna needs to be. For another thing, the rate at which you can send information depends on the bandwidth (i.e., the range of frequencies you're using). There just isn't much bandwidth at the low-frequency end.