As for your second question, I don't see what the trouble is. It looks like a straightforward generalization of Gauss' Law.
In our 3D universe, Gauss' Law for gravitation says that for every closed surface:
Surface Integral (E dot dA) = M(in)*k
where E (at a particular point on the surface) is the gravitational field, dA (at a particular point on the surface) is a vector perpendicular to the surface, pointing outwards, with infinitessimal length, M(in) is the mass enclosed by the surface and k is a universal constant that depends only on the units that we're using.
Using the same logic used to derive Gauss' Law in 3D, it shouldn't be hard to say that in 4D, we would have:
Hypersurface Integral (E dot dV) = M(in)*k
Now, suppose our distribution of mass is radially symmetric (and finite in size). Then it makes sense to choose a (4D) hypersphere as our Gaussian surface. Let's choose our hypersphere so that it encloses the entire distribution of mass, so that M(in) = M, the total mass of the distribution.
Since our distribution of mass is radially symmetric, we would expect that the magnitude of the gravitational field is constant over our hypersphere. Thus, we can pull E out of the integral, and we have:
|E| * Hypersurface Integral (dV) = Mk
But the Hypersurface Integral of dV is just the sum over all the infinitessimal surface volumes. This sum is just the total surface volume of the hypersphere, which according to
http://mathworld.wolfram.com/Hypersphere.html is 2(pi^2)(r^3), where r is the radius of the hypersphere.
So:
|E| * 2(pi^2)(r^3) = Mk
|E| = M * k/2(pi^2)(r^3)
If we define G = k/2(pi^2), that equation becomes:
|E| = MG/r^3
Since we normally expect gravity to be an attractive force, the gravitational field should point towards the origin. So finally, we get:
E = -GM/r^3 * r(hat)
where r(hat) is a unit vector that points fom the origin to the point in question.
Notice that this does not depend on the distribution itself (so long as the distribution is radially symmetric, and we're outside the entire distribution).
Since a solid glome is radially symmetric, yes, we can treat it as a point mass (which is also radially symmetric), so long as we're talking about points outside the glome.
And for your fourth and final question; since I don't even know enough to understand the proof that standard 4D orbits are unstabe; I'm afraid you'll have to ask someone with more mathematical knowledge.