Why does a phonon obey the Bose statistic? Could somebody please explain why the phonon must be a Boson (strictly speaking, it must obey the Bose statistic) regardless what it is composed of?  (As I have heard, the lattice vibration of both Bosonic and Fermionic system obey Bose statistics.)
 A: Lattice vibrations are mechanical ("sound") waves, so amount to collections of coupled harmonic oscillators. Their normal modes amount to individual oscillators with the usual continuous spectrum, and, upon ("second") quantization, discrete spectrum, linear in the integer oscillator excitation number, and unbounded above--until the lattice is torn apart. It does not matter what the constituents of the lattice are, fermions, bosons, or whatever. The positions of these constituents oscillate, and it is these  motions which are quantized.  
These quantum oscillators then are bosons.  
If, instead, for some recondite reason, the spectrum of these modes only had occupation numbers 0 or 1, you'd call them fermions, but, as already hinted in the comments, it's hard to conceive of a classical mechanical collective motion that would quantize to fermions with such a ferociously limited spectrum.  A classical limit is often associated with large occupation numbers, unavailable here, for each mode. 
I have to confess I am out of my depth with topological fermions, however, and to what extent these exotic collective excitations are fermionic. Perhaps a condensed matter person could bring expertise to bear.
