Black holes are one of highest density objects in the universe. Because of this high density(m/v) more matter is highly compressed in a very small volume( i am guessing that molecules cant freely move in black hole thus forming a perfect crystals). my question is whether entropy($\bigtriangleup S$) of the black hole can be considered as zero or not? Then what is the free energy of the blackhole, whether it will like this $\bigtriangleup G=\bigtriangleup H$.
You have a misunderstanding about black holes: objects are not trapped and immobile at the center, they are smeared out nonlocally on the horizon, and the horizon jiggles in a random way which is thermal. Black hole entropy is proportional to the area, and the horizon surface area of a black hole with finite mass in our universe is never zero.
The argument that black holes have a high entropy (infinite in the classical limit) is given by Bekenstein--- if you throw an equal mass into a black hole, the final state does not care what mass you threw in. If you throw cold matter, you get the same final state as if you throw hot matter. So the final state must have a huge entropy, equal to the maximum entropy of what you can throw in. This was the source of the black hole advances of the last 30 years, that led to AdS/CFT.
In our 3+1 dimensioal universe with no long-range scalar fields and only electric charge, you can cool a black hole down to absolute zero (by spinning it or charging it up), but the result is a large entropy state, because the horizon area doesn't go to zero in this limit. But in higher dimensions, with a dilaton scalar field and p-form charge, the 3-brane of type IIB string theory has zero entropy in the extremal limit. This is why it is described by a gauge theory (a regular quantum field theory) whose zero temperature state has zero entropy too.