Can only two equal EM-wavelength interfere? If so why is that? When there are two waves of light with the same wavelength they can constructively or destructively interfere with each other. But can for example a wave of 532nm interfere with a wave of 533nm. How exactly must they be the same and why can't (if so?) different wavelengths join together?
 A: They do interfere, though the effect is not quite the same as the effect we think of when we think of interference between light waves.
In the single frequency case, there are distances from the two wave sources for which the wave from one source perfectly cancels out the wave from another source.  These are called nulls.  If we vary the phase between the two sources, we can shift those nulls from one side to another.
If you have two different frequencies, you can treat that as though you had two sources of the same frequency, but they're constantly phase shifting apart.  If I have two sine waves $\sin(\omega_1t)$ and $\sin(\omega_2t)$, I can simply rearrange the second one to be $\sin(\omega_1 t + (\omega_2-\omega_1)t)$.  Now this should be interesting because we now have two sine waves of the same frequency, but they have a continuously changing phase difference of $(\omega_2-\omega_1)t$.
These are called beats.  In music, we actually use them to tune instruments.  If two instruments are slightly out of tune with each other, you'll hear "beats" as the sound pulsates.  Sometimes that phase shift will cause constructive interference, other times it will cause destructive.
The same thing happens with your laser, only there's an issue.  If you run the numbers, a nm in wavelength difference is something like 10^17 Hz!  The beats happen so fast here that you can't see them at all, so it appears as though there is no interference at all.
A: They can't constructively or destructively interfere because the two sinusoids go in and out of phase with each other. 
First of all, it has more to do with frequency than wavelength. Obviously, those two are connected by the speed of propagation. If we have two waves of the form
$f_1(x,t)=\cos(\omega_1 t - k_1 x + \phi_1)$
$f_2(x,t)=\cos(\omega_2 t - k_2 x + \phi_2)$
where $\omega_i$ is the frequency, then $k_i = \omega_i v = 2 \pi / \lambda_i$. Here, $v$ is the speed of propagation, $\lambda_i$ is the wavelength, and $k_i$ is the propagation constant. $\phi_i$ is some phase offset. At some specific position $x=x_0$, these waves take the form.
$f_1(x_0,t)=\cos(\omega_1 t + \tilde{\phi}_1)$
$f_2(x_0,t)=\cos(\omega_2 t + \tilde{\phi}_2)$
Now obviously if we sum these two together with $\omega_1 = \omega_2$, then if $\tilde{\phi}_1=\tilde{\phi}_2$ we get constructive interference, and if $\tilde{\phi}_1=\tilde{\phi}_2+\pi$ we get destructive interference. 
If the frequencies are different, we can use a trig identity to show that
$\begin{array}{lll}
f_1+f_2 & = & \cos(\omega_1 t + \tilde{\phi}_1) + \cos(\omega_2 t + \tilde{\phi}_2) \\
& = & 2 \cos(\frac{\omega_1 + \omega_2 }{2}t + \frac{\tilde{\phi}_1 + \tilde{\phi}_2}{2}) \cos(\frac{\omega_1 - \omega_2 }{2}t + \frac{\tilde{\phi}_1 - \tilde{\phi}_2}{2})
\end{array}$
This can be interpreted as a sinusoid of the average of the frequencies "beating" inside an envelope of half the difference in frequencies. 

Where this envelope is zero, we can understand this as a temporary destructive interference of sorts, and where the envelope is its maximum at 2, we can understand it as a temporary constructive interference. 
However, rigorously it is neither constructive nor destructive interference. 
A: Yes they can interfere. 
All interference phenomena are based on the principle of superposition. For electromagnetic fields this principle states that the total electric field is given by the sum of the individual electric fields originating from different sources. 
If we have two sources whith the same wavelength whose signals come together on a screen or a detector, the detector will measure a high intensity when the two waves are in phase and thus interfere constructively, while it will measure a low intensity when the waves are out of phase and interfere destructively. Whether the waves are in or out of phase in this case only depends on the difference in path length between the two waves. 
One consequence is that if we would move both sources backwards by the same distance, this would not influence the interference pattern.
Now if the two waves do not have the same wavelength, the situation is slightly more complicated. The superposition principle still holds, thus when the two waves both are in phase at the detector, we get constructive interference and when they are out of phase destructive interference. 
However, this time the phase difference between the two waves does not just depend on the difference in path length, but also on the total path length. Even when there is no path length difference the two waves will start to run out of phase after a certain distance. For small wavelength differences $\Delta \lambda$, the signals will have obained a phase difference of $\pi$ After $N=\frac{\lambda}{2\Delta\lambda}$ wavelengths. After twice this distance they will run in phase again. 
In this new situation moving both sources backwards by the same distance does alter the interference signal, because of the increase in the phase difference. 
If the distances in your experiment are much smaller than N wavelengths, you will not notice the effect of the wavelenght difference. 
A similar effect occurs when there is some spread in the wavelength of a single source. Google "coherence length" for more information.
A: Waves of different wavelength/frequency interfere intermittently, which produces periodic modulation called beat frequency.  The beat frequency corresponds to the difference between the frequencies.  It is what musicians are listening for as they tune their instruments.
 It's easiest to visualize this by thinking about two waves that are only slightly different. They interfere for a few waves until they are out of phase and then wander back into phase again.
