Presentations of simple groups Finite simple groups (non-abelian) can generated by two elements.
Let $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ be a finite simple group (non-abelian), and $\langle x,y|x^p=y^q=(xy)^r=1,...\rangle$  be another presentation of $G$. (Here, "..." means possibly more relations). 
1) If $(1/l)+(1/m)+(1/n)<1$, then does it imply that $(1/p)+(1/q)+(1/r)<1$? 
2) If $(1/l)+(1/m)+(1/n)=1$, then does it imply that $(1/p)+(1/q)+(1/r)=1$? 
3) If $(1/l)+(1/m)+(1/n)>1$, then does it imply that $(1/p)+(1/q)+(1/r)>1$? 
(For a group $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ to be a group of symmetries of a compact orientable surface, there are some restrictions on the integers $l,m,n$ and genus of surface, due to Riemann-Hurwitz relation. If the integers (l,m,n) satisfy some conditions as above, then it will allow us to consider less number of presentations of finite simple groups to check for possibility of action of the group on the surface).
 A: This is an attempt to explain Jack's answer geometrically, and to argue that $A_5$ is the only possible interesting (ie non-abelian) example.
We are interested in actions of finite simple groups on surfaces $\Sigma$.  We are looking for simple groups $G$ that act on surfaces of different curvatures.  The quotient $\Sigma$ is an orbifold of the same curvature as $\Sigma$, so we can just appeal to the classification of such orbifolds. (See, for instance, Peter Scott's article The geometries of 3-manifolds for a nice introduction to 2-dimensional orbifolds.)
There are only two infinite families of spherical orbifolds, having cyclic and dihedral fundamental groups.  As these have no non-abelian simple quotients, we discard them. The remaining orbifolds correspond to the symmetries of the Platonic solids.  The only interesting simple quotient we see here is $A_5$.
Famously there are seventeen Euclidean 2-orbifolds, and each has a free abelian subgroup of finite index, which we may as well quotient by.  Now, I don't have a classification of these quotients to hand (the wikipedia article on Wallpaper Groups uses Conway notation, with which I'm not comfortable), but I'm willing to bet that, once again, no interesting simple quotients arise.
Finally, we are left with the hyperbolic case.  It's a theorem that every finite group can be realised as the symmetries of a closed hyperbolic surface, so of course $A_5$ is no exception.  Jack's calculation appears to show that $A_5$ acts on a small enough hyperbolic surface $\Sigma$ that the quotient is covered by (and hence must be, I suppose) a triangular orbifold.
A: The only groups where 1/l+1/m+1/n is greater than one are the cyclic groups, the dihedral groups, and the groups A(4), S(4), and A(5). Therefore, A(5) is the only simple groups where 1/l+1/m+1/n>1. As mentioned above, it is also a group such that 1/p+1/q+1/r<1, so it is a counterexample to (1) and (3).
All groups such that 1/l+1/m+1/n=1 are solvable, so therefore (2) is vacuously true.
Therefore your statement can be made into the stronger (and more accurate) statement: The only simple group with presentation $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$, when 1/l+1/m+1/n>1 is the group A(5)
