Helium Nucleus as boson I am rather perplexed with this fact that though Helium Nucleus is a boson, the particles insides it: protons and neutrons are essentially fermions. How the nucleus which is made by fermions can be a boson ? Does spin has a role in it?
 A: The amount of detail you can see in a system depends on the energy scale you use. Although a helium nucleus is indeed a composite object the spacing of it's energy levels is much, much greater than kT at the kind of temperatures we use for experiments. That means it behaves as a single object, and its spin is simply the total spin of the ground state. You would need to be using energies of tens of MeV before the helium nucleus started behaving as a composite object.
A: I strongly recommend reading the chapter 4. Identical Particles of The Feynman Lectures on Physics, Volume III. All three volumes have been made available for free, links are pasted at the end of the post.
Quotes from The Feynman Lectures on Physics, Volume III, chapter 4:
Here's two relevant small quotes from that chapter: 

So the rule is that composite objects, in circumstances in which the
  composite object can be considered as a single object, behave like
  Fermi particles or Bose particles, depending on whether they contain
  an odd number or an even number of Fermi particles.
All the elementary Fermi particles we have mentioned—such as the
  electron, the proton, the neutron, and so on—have a spin $j=1/2$. If
  several such Fermi particles are put together to form a composite
  object, the resulting spin may be either integral or half-integral.
  For example, the common isotope of helium, He$^4$, which has two
  neutrons and two protons, has a spin of zero, whereas Li$^7$,
  which has three protons and four neutrons, has a spin of $3/2$. We
  will learn later the rules for compounding angular momentum, and will
  just mention now that every composite object which has a half-integral
  spin imitates a Fermi particle, whereas every composite object with an
  integral spin imitates a Bose particle.

Near the end of that chapter, he briefly discusses the behaviour of $\alpha$ particles in liquid Helium, I quote again:

Liquid helium has at low temperatures many odd properties which we
  cannot unfortunately take the time to describe in detail right now,
  but many of them arise from the fact that a helium atom is a Bose
  particle. One of the things is that liquid helium flows without any
  viscous resistance. It is, in fact, the ideal “dry” water we have been
  talking about in one of the earlier chapters—provided that the
  velocities are low enough. The reason is the following. In order for a
  liquid to have viscosity, there must be internal energy losses; there
  must be some way for one part of the liquid to have a motion that is
  different from that of the rest of the liquid. This means that it must
  be possible to knock some of the atoms into states that are different
  from the states occupied by other atoms. But at sufficiently low
  temperatures, when the thermal motions get very small, all the atoms
  try to get into the same condition. So, if some of them are moving
  along, then all the atoms try to move together in the same state.
  There is a kind of rigidity to the motion, and it is hard to break the
  motion up into irregular patterns of turbulence, as would happen, for
  example, with independent particles. So in a liquid of Bose particles,
  there is a strong tendency for all the atoms to go into the same
  state—which is represented by the $\sqrt{n+1}$ factor we found earlier.
  (For a bottle of liquid helium $n$ is, of course, a very large number!)
  This cooperative motion does not happen at high temperatures, because
  then there is sufficient thermal energy to put the various atoms into
  various different higher states. But at a sufficiently low temperature
  there suddenly comes a moment in which all the helium atoms try to go
  into the same state. The helium becomes a superfluid. Incidentally,
  this phenomenon only appears with the isotope of helium which has
  atomic weight $4$. For the helium isotope of atomic weight $3$, the
  individual atoms are Fermi particles, and the liquid is a normal
  fluid. Since superfluidity occurs only with He$^4$, it is evidently a
  quantum mechanical effect—due to the Bose nature of the $\alpha$-particle.


Relevant links:


*

*The Feynman Lectures on Physics

*The Feynman Lectures on Physics, Volume III, Quantum Mechanics

*Chapter 4. Identical Particles

*For a better understanding of spin composition, see this excellent answer by Edward Hughes.

