Pure hydrogen star 
*

*What is the smallest mass of pure hydrogen that can ignite fusion? 

*That is can population III stars have tiny masses? 

*How would such stars develop? 

*How long would such a star last?

 A: First, note that Population III stars are expected to be massive, not tiny, with masses upwards of $10^6\,M_\odot$. The reason for this is due to the Jeans criteria, where the mass follows
$$
M_J\propto T^{3/2}
$$
In the early universe (~ 1 Myr), the temperature was around 10,000 K; so in a pure-hydrogen environment, no cloud with a mass less than about a million solar masses could collapse to form a star.
However, some research has shown that $H_2$ could exist inside hydrogen clouds (UV light from other Pop III stars would have dissociated the $H_2$, so it must be buried), then the effective temperature would be a few hundred kelvin, dropping the masses in the $100\,M_\odot$ range, cf. this presentation (PDF).
I do not believe that we should expect the minimum mass of a Pop III star to be much different from a Pop II star's minimum mass of $\approx0.07\,M_\odot$. That said, much of the work I've seen on Pop III stars look in the range of $40\leq M_\star/M_\odot\leq1000$ (granted, much of the papers I looked at were interested in the supernova explosions of such massive stars). 
If these low-mass Pop III stars existed, they might actually still be around today and could be observed with James Webb Space Telescope. On the other hand, massive stars, like rock stars, burn bright & die young; massive Pop III stars aren't much different, with lifespans of a few million years.
A: As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium.
Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by electron degeneracy pressure. They will never become hot enough to fuse anything beyond deuterium, hydrogen and lithium. Instead they will slowly fuse all their hydrogen, but because they are fully convective they will be able to resupply the core with fresh fuel until it is all gone. This scenario is discussed by Laughlin, Bodenheimer & Adams (1997) in the context of a solar composition - they suggest this phase takes around $10^{13}$ years for a star just above the minimum mass for hydrogen burning. Once the fuel is exhausted, the star will continue to cool and will contract a little more until its electrons become completely degenerate. At which point it can continue to cool, but now at constant radius since the pressure becomes independent of temperature. Such a star (a helium white dwarf) will just sit and cool forever, until their protons decay, or something else externally happens to them or the universe.
The answer to the first part must come from theoretical models. The canonical work by Chabrier & Baraffe (1997) find a "minimum mass" for hydrogen burning (which they define as that mass where hydrogen burning can supply the stars luminosity after 1 billion years) is about $0.072M_{\odot}$ for a star with solar metallicity, but rises to about $0.083M_{\odot}$ for stars which have less than a tenth of the solar metallicity. Earlier work by Baraffe et al. (1995) determined a limit of $0.09M_{\odot}$ for a metallicity 30 times less than the Sun. The review of Burrows et al. (2001) says that the minimum mass for a "zero metallicity" population III star is $0.092M_{\odot}$. This limit originates from modelling work performed by Saumon et al. (1994). 
I am not aware of anything more recent, nor of anything more recent that tackles the entirely hypothetical question of a pure hydrogen star. However, one could wave ones hands and say that the virial theorem tells us that the central temperature is proportional to $\mu M/R$, where $\mu$ is the mean atomic mass in the gas. If hydrogen burning starts at a fixed central temperature, then the mass at which that occurs will be proportional to $\mu^{-1}$. If we remove the helium from the mixture, then $\mu$ goes down from around 0.6 to 0.5. So this very crude argument (it implicitly assumes a perfect gas and that the envelope opacity and hence radius is not changed significantly) might suggest the minimum mass for hydrogen burning rises to about $0.6\times 0.092/0.5 = 0.11M_{\odot}$. 
Further edit: Your revised question now asks about fusion, rather than hydrogen fusion. Well, deuterium burning can occur at much lower masses - probably about 15 Jupiter masses.
A: Actually, there are papers that argue that deuterium fusion can appear in planets as small as Jupiter. http://arxiv.org/abs/1506.03793
The argument is that although Deuterium fusion is thought to happen in the order of magnitude of 10 jupiter masses, the fusion of Deuterium can be facilitated by electrons "screening" the protons of Deuterium. If a Deterium nucleus "feels" less the positive charge of another deuterium nucleus, there is more probability the nuclei can tunnel through the repulsion caused by the positive charges and thus trigger fusion. 
