So after reading up some more research papers and the concepts of most people working in this field here and there, I've reached an answer I'm satisfied with. If someone disagrees or thinks there is a better or more correct way to think the problem do share your ideas.
1) First to answer my point 1. I've noticed that most di-electric surfaces' Index of refraction remains fairly constant over the visible spectrum. This means the Fresnel Reflectances remain constant and are quite low over the visible spectrum for normal incidence. For angles near 90 degrees, even rough di-electric surfaces start reflecting the incoming light specularly. You can see this data here
However in the case of metals the IOR or more specifically the Fresnel Reflectances take on the exact same value as the color (in case of gold and copper). From this I conclude that,
For metals where Fresnel Reflectances (FR) vary greatly over the visible spectrum, FR is linked and is just a consequence of the surface of metals absorbing photons of certain frequencies. This means for Gold, the electrons at the surface absorb the blue-end of the spectrum, the remaining golden colored light either gets reflected or refracted. Thus the Fresnel Reflectance is golden in color which we can also call the "Specular Color".
Note that the absorption depends on whether there are electrons available. It's possible for some blue photons to refract or reflect if there aren't much electrons for absorbing to begin with.
In-case of di-electrics the FR don't change much over the visible spectrum. However we can see color. The possible explanation is that the color in case of di-electrics comes from within the surface. That is, there are no such electrons present at the "interface" that absorb certain frequencies. Else we would have seen different FR for different wavelengths. The electrons absorbing wavelengths of certain frequencies come from within the surface, like pigments lying inside a liquid. Hence for di-electrics FR is not really linked to the concept of electrons absorbing certain frequencies imo.
This makes sense to me as colored metals (gold, copper) tint the reflections with their color. However certain di-electrics don't such as a polished marble floor. The reason for that again is that in case of metals the cause of color i.e. absorption by electrons occurs directly on the surface. The reflected light thus has a different composition for R,G,B giving us the tint. However for di-electrics the cause of color lies beneath the surface. Hence the reflection is non-tinted.
Thus simply stating for metals FR is the same as the albedo/color of the surface. For non-metals or di-electrics FR is different from the albedo/color and just defines the amount of specular reflection.
2) The proper way would surely be to use a BRDF with both diffuse and specular components since diffuse surfaces reflect greatly at near-grazing angles.
3) Still don't know about this. It might be baked in or not. What I do know is that the diffuse component should surely get lesser and specular component larger as angle increases to 90.
Another interesting point to note is that the basic Fresnel Equations apply for transparent and Flat surfaces. For metals we have complex IOR making the equations a little more complex. Sebastian Lagarde did a comparison here. Fresnel equations don't apply to rough surfaces but for microfacet models where we assume each facet to be perfectly specular we can maybe. Either way they are a good way to approximate the behaviour for increasing reflection near 90 degrees.