Probability that distance of two random points within a sphere is less than a constant

How are you at three-dimensional integrals?
Let the first point be at $(0,0,z)$. There are two spheres: One of radius $r$, centered at the origin, and one of radius $d$, centered at $(0,0,z)$. I would let $r=1$ to remove one letter from your calculations.
The spheres' intersection is symmetric about the $z$ axis, which should help you calculate the volume of the intersection. You rotate the intersection of two semicircles about the $z$ axis.
Once you have $P(z,d)$, you need to calculate the following.
$$P(d)=\frac{\int_0^1 z^2P(z,d)dz}{\int_0^1z^2dz}$$

According to MathWorld, the density for $d$ is

$$ \frac{3d^2}{r^3}-\frac{9d^3}{4r^4}+\frac{3d^5}{16r^6}\;. $$

Thus the cumulative distribution function that you're looking for is

$$ \int_0^d\left(\frac{3d'^2}{r^3}-\frac{9d'^3}{4r^4}+\frac{3d'^5}{16r^6}\right)\mathrm dd'=\left(\frac dr\right)^3-\frac9{16}\left(\frac dr\right)^4+\frac1{32}\left(\frac dr\right)^6\;. $$

Here's a plot.