Suppose the cdf of your distribution is F, whose inverse I will call G to avoid cumbersome notation. Then your quantity is the limit of 1/(n-1) [G(1/n) + G(2/n) + ... + G((n-1)/n)], which is one Riemann sum on the way to the integral of G. An easy change of variable shows that this integral equals the mean (or doesn't exist if the mean doesn't). So that resolves question 1.