As a general rule there are three mechanisms by which molecules absorb light:
In solids you don't often get rotational spectra because the molecules usually aren't free to move without interacting with the lattice, so you tend to get electronic transitions in the uv and vibrational transitions in the IR. It's probably not coincidence that there is frequently no absorption at visible wavelengths because we wouldn't have evolved eyes if there was.
In isolated molecules you get nice sharp vibrational transitions (with rotational structure as well) but in solids the interaction with the lattice tends to broaden out the absorption lines. You'll find numerous articles on IR spectroscopy of silica glass, for example this one though it's behind a paywall. As DumpsterDoofus comments, you get absorption due to hydroxyl and/or water, but you also get absorption due to various stretching modes of the Si-O-Si lattice.
In microwave ovens what matters is how much energy the radiation carries and how that energy is absorbed by the food. Visible light and IR are rapidly absorbed by most foods, so they would only heat the outer layer of the food. You'd get food with the outside carbonised and the inside raw.
Microwaves are far less strongly absorbed by foods, so they penetrate deep into the food and can heat the interior. Even so large objects won't be heated throughout, which is one reason why microwave cooking instructions frequently advise a multi stage process of heating, letting the food stand then heating a final time.
Microwave ovens often include IR heating as well as the microwave heating. This is done so you get food with a browned exterior and heated throughout.
The answers to Why do microwave ovens use radiation with such long wavelength? give a nice discussion of why the exact wavelength used was chosen. The frequencies commonly used in microwave ovens are 2.45 GHz (12 cm) for home ovens and 915 MHz (38 cm) for industrial overs. Much higher frequencies are not used due to the cost of the magnetron, while much lower frequencies would not work because the wavelengths would be too big to allow a half wavelength to fit in the oven.
Finally, you say:
Why do we use microwaves in microwave oven when infrared and visible light are much hotter and how do microwaves cook food when they are cooler than visible light and others.
But this is a slight misunderstanding. The wavelength of light emitted is indeed related to the temperature of the source, but light itself doesn't really have a temperature in the sense that matter does. Light transfers energy, and if this energy is absorbed it will heat the food. However the amount of heating is just related to the intensity of the EM radiation and the abosrption cross section. The wavelength makes a difference only insofar as it affects the absorption cross section.
Best Answer
Refraction of light in water droplets, leading to the formation of rainbows, is not limited to the visible range.
Experimental evidence, compelling due to its simplicity, is shown in the following images taken by University of College London Earth Sciences professor Dominic Fortes. Check the alignment of the rainbow with respect to the trees in each of the pictures. The UV band lies to the left of the visible band, while IR is found to be shifted to the right.
The spectral limits in a rainbow can be explained more technical by looking at the refractive index dispersion of water vapor, which can e.g. be found at refractiveindex.info. The UV, visible and near IR range lie in the wavelength region between 0.2 and 2.85 µm. The change in refractive index with respect to the wavelength leads to differing refraction angles and therefore a separation of the colors, as we know it from experience. Basically, this concept could also be extended to further wavelength ranges. Although the resonance around 2.9 µm leads to higher refractive indices for longer wavelengths again. Therefore light with a wavelength of e.g. 4.3 µm would overlay with light at 0.4 µm (both with a refractive index of 1.34). Yet, this is again only half the truth. If you look at the transmittance curve (further down on the same page), you can see that wavelengths longer than 1.8 µm are absorbed by water vapor. Therefore this is the realistic long wavlength end for rainbows. I assume similar arguments could be found for the short wavelength end, but I can't find experimental data.