Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride
I just wanted to add that it's up to my knowledge that $\ce{LiAlH4 /
H^{−}}$ can't reduce the double bond of Benzene, because when styrene
  reacts with $\ce{LiAlH4 / H^{−}}$ it gives ethyl benzene.


That answer is as good as any for the first part of your question.  It demonstrates that LAH will react preferentially with aliphatic pi bonds rather than the aromatic ring.  Similarly there are aliphatic pi bonds in $\ce{Ar - N^+\equiv N}$, and so the LAH will reduce the diazonium group rather than displacing it.

For the hypophosphorous acid reduction, the mechanism you found has a problem.  The principle of microscopic reversibility argues against three components coming off at once in what is shown as the first step (it would be equivalent to three independent bodies colliding simultaneously).  The first step may be correct but it likely has more than one component to it.