I also found this surprising when first introduced.
The light has no memory. When it passes through the second polariser, there is no information whatsoever of its previous polarisation. It could have been whatever!
When light goes through a polariser at $30º$, it gets tilted at the cost of some lost intensity. In other words, it gets projected on the polariser axis, that is rotated a bit with respect to the light polarisation vector. This new light then goes on to the next one, and gets tilted again.
You get the picture. You can use this idea to arbitrarily change the polarisation angle of light by putting many of them in series. The more steps you do, the more polarisers you use, and thus, the smaller the steps, more light will come out on the other side (in reality, you always loose some light because they are not completely transparent, so you don't want to use many of them).
$I_0$ is the intensity of light before it hits a polariser the original intensity of the beam, so called. You need it because you need to compare it to the intensity after it exits the polariser so that you can calculate your fraction of incident intensity.
this fraction requested by the problem is $I\over I_0$, but $I$ refers to intensity of light exiting the third filter (Malus' law only describes intensity in and intensity out for one filter). you would have to use Malus' Law, $I \over I_0$ $= \cos^2(\theta)$ three times. also notice the subtle rearrangement I made to make the left hand side refer to fractional intensity. this will be made clear in abit..
If I was working on the problem I would use $I_0$, $I_1$, $I_2$, and $I_3$ to refer to the original intensity, intensity exiting the first filter, intensity exiting the second filter, and intensity exiting the third filter, respectively. the final answer to the problem would then be $I_3\over I_0$ using my chosen notation...
The solution is easy to obtain once you know that
$I_1 \over I_0$ = $1 \over 2$ , this is a special case to Malus' Law when $I_0$ is unpolarised
$I_2 \over I_1$ = $ \cos^2(55-10)$
$I_3 \over I_2$ = $ \cos^2(85-55)$
Best Answer
Malus's law is about the effect of a polariser on polarised light. You've clearly read a badly written version of it. What your author likely meant to say was:
One begins with unpolarised light;
The first polariser quells the unaligned component of the unpolarised light and outputs polarised light (with half the input's intensity). This polarised output has intensity $I_0$ in your notation;
Of the polarised output from the first polariser, the second polariser lets through a fraction $(\cos\theta)^2$ where $\theta$ is the angle between the axes of the polarisers. So I say again: $I_0$ is the intensity of the polarised input to the second polariser, not the intensity of the unpolarised input to the system of two polarisers. With this proviso, the output intensity is $I_0\,(\cos\theta)^2$.
In Answer to:
Depolarised light is actually quite a subtle and tricky concept: I discuss ways of dealing with it in my answers here and here. You can think of it intuitively: without a preference for polarisation, perfectly depolarised light must dump half its energy into a polariser: you can take this as a kind of "definition" of depolarised light if you like. The quantum description is much simpler than the classical, so I reproduce it here. We imagine the source producing photons each in pure polarisation states but with random direction. That is, each photon's polarisation axis makes some random, uniformly distributed angle with the polariser's axis. Its pobability of absorption is therefore $(\cos\theta)^2$. So now we simply average this quantity given a uniformly distributed angle:
$$\left<(\cos\theta)^2\right> = \frac{1}{2\pi}\int_0^{2\pi} (\cos\theta)^2\,\mathrm{d}\theta = \frac{1}{2}$$
to find the overall probability of passage through the polariser, and thus the proportion of photons that make it through. For more info on what polarisation actually means for a single photon, see my answer here.