Why the electrons in a insulator fill up the valance band exactly? The classic picture of the band structure for an insulator is a filled valance band below $E_F$ and empty conduction band above $E_F$.
This picture seems to me that the electron density is fixed. For example, at $T=0$, $n=\int_0^{E_m}  d\epsilon D(\epsilon) $, where $D(\epsilon) $ is density of state and $E_m$ is the maximum energy of the nearest valance band.
At zero $T$ why $E_F$ is between two bands instead of $E_F=E_m$.
If such insulator is doped with more electron. Does the extra electron fill the valance band and $E_F$ increases?
 A: To answer the second question: adding only electrons would mean charging the insulator. Since all the states in the valence band are occupied (by definition, otherwise we would call it a conductor) to these electrons can go only into the conduction band.
Doping usually means adding atom that can supply or accept electrons, called respectively donor impurities and acceptor impurities. Adding atom however means adding additional states. Unlike the bands, these states are descrete. Even if there many impurities, they are usually disordered, and do not form a band of their own.
In practice one often uses donor impurities that would have their levels just below the conduction band, so even small temperature would make these electrons excited to the condition band, making the material conducting - this is called n-type semiconductor. Similarly, the acceptor impurities typically have energies just above the valence band, so that electrons from the valence band can jump into these levels, leaving behind holes, and resulting in an p-type semiconductor.
Note that the position of the Fermi level changes, when a semiconductor is doped, since at zero temperature the Fermi level should be above all the occupied energy levels, and below all the empty ones.
