Interpretation of Fermi-Dirac statistics

So the fermi-dirac distribution function has the form $$nleft(Eright)=frac{1}{exp left(frac{E-E_F}{k_bT}right)+1}$$ So for $E=E_F$ you can see that $nleft(E_Fright)=frac{1}{2}$ regardless of the value of $T$ so I am not entirely sure what you mean by the first statement in your question.

If you plot the FD distribution function at different energies you will see that the graph "softens" . (Image taken from wikipedia http://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics)

If you consider the function at zero temperature you have no states filled above the Fermi level and every state filled below the Fermi level. As you increase the temperature occupied states which are within $approx k_bT$ of the Fermi level can be thermally excited to empty states above the Fermi level. So you can think of it as states from energies roughly $E_F-k_bTrightarrow E_F+k_bT$ which are changing their occupation. So you could imagine that the states at the two extremes are changing their occupation very rarely i.e. the one at $E_F+k_bT$ is empty most of the time and the one $E_F-k_bT$ is mostly filled.

For the states in around the Fermi energy, they change their occupation quite regularly due to thermal excitations. The state at exactly $E_F$ is filled half the time from the form of the distribution function.

As for the semiconductors, I'm no expert on condensed matter theory but I think you could probably think of it this way. The n-type semiconductor has dopants which produce more electrons to conduct. So if you think of the material originally having $N$ electrons now we have some $N+n$ electrons to conduct, which now have to fill up more states in the material pushing the Fermi level higher.

A p-type semiconductor has dopants which remove the number of mobile electrons which can conduct current, so the converse argument applies we now have $N-n$ electrons conducting.

(I wouldn't quote me on the last bit)

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I have actually been a little lax and used a low temperature version of the FD distribution function I should be using something like $$nleft(Eright)=frac{1}{exp left(frac{E-mu}{k_bT}right)+1}$$ where $mu$ is the electrochemical potential and is related to the Fermi energy by $$mu=E_Fleft(1-frac{1}{3}left(frac{pi k_bT}{2E_F}right)^2right)$$ so that might relate to your first point. I'm just used to dealing with systems were $E_F>>k_bT$ so I didn't make the distinction.