Questions about subnuclear decay: helicity and parity. I read that the process (where the last is the antineutrino of electron): $$ \pi^{-} \Rightarrow  e^{-} + \hat{\nu}$$  
is disadvantage because of spins and parity argumentations, but why? In the 99% of the case the pion decay into muon and antineutrino of muon, I don't understand the reasons.
In weak interactions like that only left fermions and right antifermions interacts; this mean that the electron spin is in the opposite direction of the motion, right?
Second example (that I don't understand): in the following interaction:
$$ \hat\nu_{\mu} + e^{-} \Rightarrow \mu^+ + \nu_e$$ 
the electron neutrino, in system of the center of mass, is products in isotropy way or in the opposite directions compare to the electron? 
 A: The pion is a pseudoscalar, with spin-parity $J^\pi=0^-$.
In decay of a pion at rest the two decay products have equal and opposite momentum.
Choose a coordinate system so that the electron goes to the left, and the antineutrino goes to the right:
e <--------- π -----------> ν

The weak interaction, as you say, involves particles with left-handed chirality and antiparticles with right-handed chirality.  For massless particles, chirality is the same as helicity, while for massive particles the two phenomena are independent.  So if the electron and neutrino were both massless, the antineutrino spin would point parallel to its momentum (right-handed), which is to the right, and the electron spin would point antiparallel to its momentum (left-handed), which is also to the right.  If the electron and antineutrino spins are parallel you cannot have $J=0$;
a $0 \to 1$ transition violates conservation of angular momentum and is forbidden.  (Orbital angular momentum doesn't help much, for reasons too complex for a parenthesis.)
Since the electron and antineutrino are more-or-less equally sharing 140 MeV of kinetic energy, the electron has relativistic factor $\gamma = E/m \approx 100$, and the approximation that it's massless is pretty good.  So the $\pi\to e\nu$ decay is approximately forbidden.  In $\pi\to\mu\nu$, on the other hand, the muon is nonrelativistic and its chirality is less strongly related to its helicity.
I don't understand your second interaction (v1 of question), which violates conservation of charge.
