Why does $8\pi/3$ appear in the equations describing cosmological expansion? What is the significance of $8\pi/3$ in the first Friedmann Equation, and in the question concerning the time independence of the Hubble Constant?
Is it the 'same' $8\pi/3$ that appears in the total cross section  formula of Thomson Scattering ?
 A: The $8\pi G$ comes from Einstein's field equation, the normalization coming from matching to Newton's law in the nonrelativistic limit (c.f. "correspondence limit" on the wiki page). I believe the $3$ probably comes from the dimension of space, but I would have the check the derivation of the Friedmann equation again to be sure (wrong! see below). The numerical factors are arbitrary since you could absorb them into a redefinition of $G$ if you like. So it's really just historical convention to make the Newtonian limit for the force law look simpler than the Friedmann equation.
EDIT: Re Thompson cross section... nope. The relative normalization of gravitational and electrodynamic constants can be chosen however you like. Once you choose the standard conventions it just falls out how it falls out, but there is no "fundamental" meaning to a numerical coincidence when you have independent dimensional constants available to adjust.
EDIT 2: I was wrong about the $3$. The factor in Friendmann's equation in $d$ spatial dimension is $$\frac{16\pi G}{d\left(d-1\right)}.$$
It does come from the dimension of space but not in the obvious way.
A: Whenever you see products of $\pi$ and the gravitational constant $G$ appear in equations (terms like $4\pi G$, $8\pi G$ or $16\pi G$), please realize that this is due to Sir Isaac failing to put a surface area rather than the square of a radius in the denominator of his gravity law. Failing to use the "corrected" gravity equation: $$F = G' \frac{m_1 m_2}{4 \pi r^2}$$ and continuing with $G$ instead of $G' = 4 \pi G$ caused a funny factor of $4\pi$ to appear in the Poisson (potential) formulation for Newtonian gravity: $$\nabla^2 \phi = -4\pi G \rho.$$ 
All of this suggests $G' = 4 \pi G$ to be the "true" gravitational constant. When used in the first Friedmann equation the factor $8 \pi G/3$ would be replaced by $\frac{2}{3}G'$.
Based on Einstein's work, others might argue that $8 \pi G$ or $16 \pi G$ are "more fundamental", but in any case it would certainly be meaningful to absorb a factor $\pi$ along with some power of two into the gravitational constant.
A: In my opinion, the $8\pi$ comes from the surface are of the fundamental (spherical) quantum singularity (or $4\pi$) multiplied by the factor 2 which compensates for the $\frac{1}{2}$ factor in $\frac{1}{2}Mv^2$ which is not needed at that most fundamental level because there is no action/reaction.
