Do Maxwell's equations predict the speed of light exactly? I know that $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ is equal to the speed of light but is this prediction accurate?  I mean is it 100 percent accurate?
 A: The wikipedia article has some great info. on the historical progression of this question.
Using Maxwell's equations, one recovers the wave equation for electric and magnetic fields. From these equations, Maxwell postulated that light can be thought of as an electromagnetic wave since the electric and magnetic fields solve the wave equation with a phase velocity of $c$. This theoretical value of the phase velocity of light waves is exactly $299,792,458 \frac{m}{s}$.

I mean is it 100 percent accurate?

Nothing is ever 100% accurate - that's an idealization of the human mind since there is always systematic errors in the measurement process (even if perfect humans are conducting the experiments). So, no, it is not 100% accurate. It's worth noting that there's a difference between accuracy and precision.
As the wiki article shows, the precision and the accuracy of the measurements of the speed of light have improved greatly with time: as better measurement techniques are used, the measurements agree with each other better and better, and any one of the measurements agree with the theoretical value of $c$ better and better.
So, although we cannot have 100% accuracy in principle, we can have such small uncertainties in the measurement that we might as well consider the value to be 100% accurate. From the wiki article,

After centuries of increasingly precise measurements, in 1975 the speed of light was known to be $299792458 m/s$ with a measurement uncertainty of 4 parts per billion.

A measurement uncertainty of 4 parts per billion is very small, i.e. 0.0000001% of uncertainty: it's like saying for every million years you live you only fudge 30 seconds of it.
Lastly, in 1983, the meter was redefined in the International System of Units (SI) as the distance traveled by light in vacuum in 1/299792458 of a second. This is justified because, as I said above, the precision of measurements of $c$ are so precise that we might as well just define $c$ to be exactly the value that they all (within certainty) agree upon - this value as it turns out is also what Maxwell's equations predict.
Thus, yes the value of $c$ is defined to be exactly that given by Maxwell's equations, but importantly this is justified because the measurements are very very very precise.
A: The answer is unequivocally no: as the quantities are currently officially defined in SI, $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ is absolutely not a prediction of the speed of light. The speed of light is defined to have a certain, exact value. That is what forms the basis for the definition of the meter. If someone came along and made a more accurate measurement of the speed of light, it would actually not change the speed of light: it would change the meter. 
In theory, this would mean that all the meter sticks in the world would need to be thrown out.  But in practice, any change would be orders of magnitude smaller than the precision of a meter stick, or any other common way of measuring distance, so it would actually not lead to real world changes at all. It's still worth emphasizing that the effect of a more accurate measurement of the speed of light would not be to change the numerical value assigned to that speed, or to any other constant, but rather to introduce a tiny (in almost all cases, too tiny to matter) bias into all the measurements which have ever been made that depend on the definition of the meter, which include not just distance, but also quantities whose units include meters in their definition, such as force, etc.
The same goes for $\mu_0$ and $\varepsilon_0$: both have defined values, and if increasing experimental accuracy were to lead to a discrepancy, say in the amount of force measured between charges or currents (doubt this is the most sensitive way to do such experiments, but it's just a hypothetical), this discrepancy would not be remedied by changing the values of $\mu_0$ and $\varepsilon_0$ (or any other official value). It would instead be done by tuning the calibration of the apparatus used to make the measurement so that it gave the desired result.  The apparatus would then presumably (eventually) become part of the new standard method for most accurately experimentally reproducing the meter, the kilogram, the ampere, or whatever.
Addition in response to comment by @garyp
As of May 2019, SI will be redefined. $c$ will still have a defined value, but the charge on the electron will change from being subject to measurement to being a defined value, in coulombs. The meter, as we already said, is and will continue to be defined by $c$, the second will continue to be defined by cesium radiation, and the kg will now be defined by specifying a value for Planck's constant. The units of $\varepsilon_0$ are $\frac{C^2 s^2}{m^3 kg}$, so are already completely defined.  Thus it will no longer be logically consistent to assign a defined value. So starting next May, the values assigned to $\mu_0$ and $\varepsilon_0$ will be subject to change on the basis of more accurate experiments. If one is changed, the other will need to change as well (along with other values, including the fine structure constant), to maintain $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$.
A: $\let\eps=\varepsilon \def\qy#1#2{#1\,\mathrm{#2}}$
I always wonder seeing how few people really understand that units
systems and physical dimensions are entirely conventional, and that as a
consequence many "universal constants" are matter of convention as
well.
So that e.g. "measuring" $\eps_0$ may get quite different meanings
according things are defined. You can't measure $\eps_0$ unless a unit
of charge is independently defined. Otherwise, how can you measure a
capacity? 
In SI as it still defined (before the new definition enters in force
next year) $\eps_0$ is a fixed constant: $\eps_0 = 1/(\mu_0 c^2)$,
with $\mu_0 = \qy{4\pi\cdot10^{-7}}{H/m}$, $c=\qy{299792458}{m/s}$.
(More exactly, the value of $\mu_0$ enters definition of unit of
current, $c$ in the one of metre.)
In the new SI there is a complete change, as now the coulomb (and
therefore the ampere) are defined in terms of the electron charge.
Then it will make sense to speak of a measurement of $\eps_0$ or
$\mu_0$. This is better explained by Ben51.

Some words about @Andrew Steane's answer. He's certainly right in
saying that our present view of electromagnetic field is much more
complex than Maxwell's theory could encompass (how could it not be so,
given 150 years in between?) After all, the same can be said for the
whole of physics...
The main difference is that Maxwell's theory is a linear one. A
consequence is that if two light beams intersect in the same region of
space, they pass each other unhindered. (Interference is no
counterexample, but I don't want to dwell explaining why.) Today we
know that it is not exactly so: e.g. photon-photon scattering exists.
But its probability is so faint that AFAIK there is only indirect evidence
of such process, no direct experimental proof.

"well, $\eps_0$ and $\mu_0$ don't exactly capture the physics of
  these fields, so the answer has to be 'no'".

Nobody could disagree, but... IMHO we should be very cautious with
statements like this when speaking to non-specialists. To
understand its actual reach it's necessary to have an adequate
understanding of how physics works. A simple "no", even though strictly
right, is at risk of being interpreted too literally. After all Andrew
also writes

Our best understanding is offered by quantum field theory and
  general relativity, and even those are probably not the whole story.

It will never be the whole story, in the whole of physics. It's the adverb "exactly" in question's title which had to be criticized. Nothing in physics is "exactly" true, but at the same time an overwhelming part of our knowledge does have wide fields where it can be safely and confidently applied. As to Maxwell's electromagnetism, think of how many electromagnetic devices are in
general use today, both between domestic walls and in sophisticated
scientific laboratories and space undertakings. It would take a post much longer than this just to touch, as an example, upon how complex and demanding is GPS system, how deeply it relies on our thorough understanding of
electromagnetic fields and their propagation. Unfortunately the very
success of projects like that drives us to take them for granted.
A: I'm not sure exactly what question you are asking, but it seems to me that the answers that referred to units and definitions of distance and time (e.g. in the SI system) may be missing what the question is. 
In the back-up question mentioned as a comment, you ask "does it agree with experiment 100%?" Well no experiment gets 100% precision. The most one can ever say is "it is consistent with all experiments that have been performed, and the best precision attained so far is ...".
If your question is: "is it the case that electromagnetic waves are fully and correctly described by classical electromagnetism?" then the answer is no, because classical electromagnetism is not quite the same as our best understanding of these fields. Our best understanding is offered by quantum field theory and general relativity, and even those are probably not the whole story.
Let me flesh this out a little. In classical electromagnetism, we can measure $\epsilon_0$ by doing static experiments with things like capacitors, and we can measure $\mu_0$ by doing experiments with things like electrical inductors. If someone says those constants have defined values, then such experiments serve as a way of relating field measurements to other properties such as current and charge. Having done all that, we can then measure the speed of electromagnetic waves in vacuum. 
Within the theoretical model offered by classical electromagnetism, the answer to your question is that, yes, electromagnetic waves in empty space do propagate at exactly $1/\sqrt{\epsilon_0 \mu_0}$, and you can, if you like, regard this as a good way to define $\epsilon_0$ once $c$ and $\mu_0$ have been given their defined values. However, the physical world does not behave exactly as this theoretical model suggests. Among other effects that complicate the picture are that bright enough light waves will interact with one another in complex ways, owing to their interaction with the Dirac field which describes electrons and positrons. Also, they will cause spacetime curvature. Even the field between the plates of a static capacitor is not nearly as simple as the classical theory suggests (though the corrections are very small in ordinary circumstances). 
In the modern understanding based on quantum field theory and general relativity, light in otherwise empty space does move at the special maximum speed, but the fields are no longer fully specified by equations involving just $\epsilon_0$ and $\mu_0$ (in addition to the fields themselves), so the answer to your question is something like "well, $\epsilon_0$ and $\mu_0$ don't exactly capture the physics of these fields, so the answer has to be 'no'".
A: In the current SI system $\epsilon_0$ is defined as $\epsilon_0=1/c^2 \mu_0$ so in the current SI system $c=1/\sqrt{\epsilon_0 \mu_0}$ is clearly exact. Furthermore, $c$, $\epsilon_0$, and $\mu_0$ are themselves all exact defined quantities with no uncertainty individually.
In the new SI system starting next year we will still have $\epsilon_0=1/c^2 \mu_0$ so in the new system $c=1/\sqrt{\epsilon_0 \mu_0}$ will still be exactly true.  However, $\epsilon_0$ and $\mu_0$ will now themselves each be uncertain quantities.  Under the new system, $\epsilon_0=e^2/2hc\alpha$ and $\mu_0=2h\alpha/ce^2$. All of these quantities are exact with the exception of the fine structure constant, $\alpha$, which has an experimental uncertainty of 0.23 parts per billion. Note that the contribution of the uncertainty in $\alpha$ is such that the uncertainties cancel out and while $\mu_0$ and $\epsilon_0$ are each individually uncertain their product is exact.
