1. Reflectance is the ratio of outgoing light to incoming light. In other words, for any beam of light striking the surface it tells you how much of that light will reflect off of it instead of getting absorbed. Since it's a ratio, only [0,1] values make sense if you're going to enforce energy conservation. You can compute it for a BRDF and a light direction by integrating the result of the BRDF over the entire hemisphere of viewing directions. So you can essentially think of that process as summing up all of the light that's reflected in all directions from a single ray of light.
2. Rspec is the reflectance of the specular BRDF.
3. By "glancing angle" they mean that the vector light source is close to parallel with the surface plane. This is consistent with the common usage of the term glancing angle in the field of optics, where it refers to an incoming ray striking a surface.
4. So as that paragraph says, they compute the directional-hemispherical reflectance of the specular BRDF with the light direction held constant at ThetaI = 0. Since the reflectance is highest when θi = 0, you know that the reflectance value represents the maximum possible reflectance value for the BRDF. So by computing the maximum reflectance and then dividing the BRDF by that value, you can be sure that the reflectance of the BRDF never exceeds 1 (as long as Cspec is <= 1).
If you want to derive this result yourself, first start with the specular BRDF. This is the right side of 7.46:
As we established earlier, we can compute directional hemispherical reflectance by integrating our BRDF about the hemisphere of possible viewing directions. We'll call this set of directions Ωv, and in spherical coordinates we'll refer to the two coordinates as Φv and θv (not to be confused with θi, which refers to our incident lighting direction). The integral we want to evaluate looks like this:
The "sinθv" term is part of the differential element of a spherical surface, which is defined as dS = r2sinθdθdϕ. In our case we're working on a unit hemisphere, so r = 1.
Now we're going to evaluate this with θi held constant as zero. In this case α is equal to θv, and so we can make that substitution. We can also pull out Cspec / π, since that part is constant:
You can verify the result of this integral using Wolfram Alpha.
EDIT: I accidentally left out the "dΩ" and "dθdϕ" from the integrals in the middle image. Please pretend that I put them in there.