Why don't waves erase out each other when looking onto a wall?
1) First let us separate colour perception from frequency. Individual frequencies have a color correspondence but the colour the human eye perceives is another story.
2) White light, such as sunlight, is composed of many frequencies.
When the impinging wave hits a wall it can be a) reflected b) absorbed c) scattered incoherently
In order for the light waves to cancel out each other or double each other the photons have to be, within the uncertainty principle, superimposed in time and space. Sometimes it happens, but the probability is small. That is one of the reasons why a reflected beam can never have the same strength as its original beam. If the frequency is the same the probability will be higher than if the frequencies come from a random palette.
This superposition can be achieved with lasers, where there is control of frequency and the beam is coherent, i.e. the phases are preserved upon reflection. A hologram is an example of superposition of same-frequency light to create a three dimensional shape by peaks and dips.
Edit: From a disappeared question the following comment is worth adding:
You can perceive all colours even if only two frequencies are shining on an object. Also in this decade, Land first discovered a two-color system for projecting the entire spectrum of hues with only two colors of projecting light (he later found more specifically that one could achieve the same effect using very narrow bands of 500 nm and 557 nm light).
I'd like to take this in three parts, and I am going to stick to a classical description, though the quantum mechanical description is in some ways easier and just as valid.
Superposition is the principle that the amplitudes due to two waves incident on the same point in space at the same time can be naively added together, but the waves do not affect each other.{#}
That last bit is important enough to write again: the waves do not affect each other. They can, in fact pass right through one another.
For the purposes below we can treat every wave as being a plane wave and thus being described by an amplitude $A$, a wave number $\vec{k}$, and a phase $\phi$. The wave number encodes both the direction of propagation (that why it's a vector) and the frequency (in the absolute value) $|k| = \frac{\omega}{c} = \frac{f}{2 \pi c}$.
This description is not complete because (a) the planar nature is only approximate and (b) even allowing for that the phase can be a function of position on the wave front; but we can safely neglect these issues.
So at any given point in space and time we have a big jumble of different waves all added together $$ \sum_{\text{directions}} \int d\omega \int d\phi A(\vec{k},\phi) e^{i(\vec{k}\cdot\vec{x} - \omega t + \phi)} $$
You eye is a camera: it has a small opening with a focusing element (lens) at the front and a light sensitive projection surface (retina) at the back. It sorts out incoming light waves two ways.
First, the lens focuses light to a particular point on the retina according to it's direction. That makes the sum over direction go away, because each patch on the retina sees only one direction (well, a very small range of directions). If the lens isn't the right shape you lose some of this and the image becomes blurry. Then you need glasses.
Secondly individual patches of the retina are sensitive to light of different frequency ranges{*}. So for each rod or cone, the integral over angular frequencies $\omega$ is reduced to quite moderate limits (human vision extends over roughly one octave in the electromagnetic spectrum).
That still leaves us with the sum over the phases, and I am going to elide this part of the problem by saying that they are roughly constant over the small dimensions of the eye's lens and time-scale of you retinal response; if that doesn't make you happy I will refer you to genneth's comment: $\langle E^2 \rangle \neq \langle E \rangle^2$.
Radios live in Fourier space:{+} They typically accept signals from a very wide set of direction, so we don't get the spacial filtering that you get with the eye. However, the electronics that back them select only a very narrow band of frequencies. In essence inspecting the incoming signals in Fourier space. And the signal you are looking for has an additional periodicity (either amplitude in AM or frequency in FM) in the audible range as well, so another round of filtering is done and you generally have a dominate remaining signal. Intentional signals have constant phases, while unintentional signals (i.e. background or noise) have random phases and they tend to cancel out.
{#} In another answer anna talks a little bit about the conditions under which this classical statement breaks down.
{*} The mechanism by which the rods and cones respond to light in inherently quantum mechanical, the means by which those signals are propagated to the brain is inherently biochemical and the post processing done by the brain is huge topic in and of itself; all of which is beyond the cope of this discussion.
{+} Traditional radios. Like the AM/FM job in your grandfather's car (since we're all 21th century people and have spiffy combined GPS-mp3 player widget or something). Ultra-wide band like your wireless base station is another matter.
You asked two questions. First, the light:
As has been pointed out, typically (there are exceptions) the colorful wall is not emitting anything. Let's take blue, for an example. White light shines on the wall. White light is composed of many different frequencies/wavelengths, which we would perceive, if they were busted out with a prism, as many different colors. The blue wall absorbs all these colors except the blue. So there are not a lot of competing wavelengths coming off the wall; the blue paint allowed the wall to absorb all the competing wavelengths of light, and reflect only blue. Another way of looking at it is that the wall is covered with a "blue-reject" absorbtion filter. We get to see the one color that the wall ISN'T.
But, there may be many competing light sources, with random orientations and intensities, illuminating the blue wall. Again, everything is absorbed except blue. The blue light we see is randomly oriented and phased. No problem. The sum of the blue light vectors from any given point will still add up to a resultant wave that is of blue frequency. Yes, if you planned it very, very carefully (e.g. used polarization equipment and lasers with controllable phase offsets) you could engineer some strange effects. Normal (non-laser) light is much more random than that.
Second, the radio:
Radios run the same way: they receive E-M energy from many different sources of random orientation, polarization, intensity, and frequency. The radio antenna acts as a band select element, which cuts out the vast majority of "junk" E-M radiation. (For instance, your car's radio antenna is tuned to pick up the waves from about 0.54MhZ (bottom end of the AM band) to 108MHz (top end of the FM band). It isn't that great for picking up microwave oven radiation at 2450.MHz, nor the 60Hz from the AC lines which power said uWave oven. Still, some of that E-M junk does get onto the antenna. Modern car radios use superhetrodyning to mix all this junk with an internally generated frequency. The result is ANOTHER spectrum of junk which is frequency shifted. The math is done perfectly so that only the radio frequencies of interest get shifted into a very narrow slot (band select filter) that the radio then processes for you to hear.
Sometimes, ESPECIALLY with AM radios, you will get fade. Sometimes this is just because of blind spots due to terrain and atmospheric bending. Sometimes it is one of the phenomena you described: multipath interference. The radio wave which reaches your antenna has followed two or more separate paths, with at least one of those paths being longer than the other(s) by odd multiples (1, 3, 5, ...) of one-half a wavelength. The waves hitting the antenna add to zero, and you hear nothing.
I hope that helps.