I could not understand the logic of this circuit, connecting an inductor to the supply what will change?
The inductor acts as a way to bias the device. It is a short circuit at DC (so you can provide a DC bias current) but is an open-circuit at the frequency of interest. If you didn't have it there in this circuit your MOSFET would have no DC bias current and thus would be off. But yes, for AC analysis you simply remove the inductor.
Considering that C1 acts like a short circuit in AC analysis why do we differentiate x and Vout ? Are not they the same node?! How would the circuit looks like when L1 becomes an open circuit?
In AC analysis x = Vout only at infinite frequencies. We sometimes make the approximation that C1 acts like a short circuit but this is not rigorously true. Sometimes you may wish to study the effect of C1 on the frequency response. But you are correct, in this case the way the question is worded, it isn't important to distinguish them.
Why there is a connection between the middle of NMOS and the ground node, and how can we connect the outer middle part of NMOS anyway? Finally, I also did not understand the "limiting factor" part of the question
A MOSFET is a 4 terminal device. In discrete MOSFET's usually the 4th terminal, called the "bulk" or "substrate" is connected to the source terminal internally so you see the MOSFET as a 3 terminal device so the answer to how we connect the outer middle part is that we don't -- the manufacturers do it for us.
Alternatively if you are in an IC technology then you DO have control over this, but most processes are what are known as NWELL processes and will have the substrate grounded for an NMOS device no matter what.