Good job, Silo, but the old “45[sup]o[/sup] for maximum range” figure depends on a flat surface and uniform gravitational field. Since we’re going a significant way around the Moon, if not all the way, neither approximation is valid. To get the maximum range (i.e., closed orbit not intersecting the surface), it’d be easiest to use a circle, or firing the bullet exactly horizontal. I’ll get back to you on the calculations.
Maybe a little less than 45 degrees (say 37.5 degrees) will get you that maximum range figure in the a-fore-mentioned situation but I’m not quite sure. So where are those calculations? 
I think I’m close to the answer give or take 20 miles.
OK, I’ve finally got some numbers. The calculations for this were kind of horrid, and would be even worse without diagrams, so if nobody minds, I’d rather not post them unless challenged. Assuming that the bullet is fired horizontally from the top of a 2 km mountain, with a muzzle speed of 914 m/s, and using the values 1738 km for the radius of the Moon, 7.3510[sup]22[/sup] kg for the mass, and 6.6710[sup]-11[/sup] m[sup]3[/sup]/kg*s[sup]2[/sup] for G, I got that it would travel .005 times the distance around the Moon before hitting, or a distance of 54.13 km (33.6 miles). After going through the calculations, I’m inclined to believe that Silo’s approximation at 45[sup]o[/sup] wasn’t so bad after all, but I’m not going to crank the numbers for an elevation of 45[sup]o[/sup], or worse yet, figure the optimum elevation, unless I’m getting a grade for it-- it took me over an hour to get what I did, and that’s the easy case.
To sum up: I think that Silo’s right, but I don’t want to test it.