Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride Why does benzenediazonium chloride get converted to benzene upon
reacting with $\ce{H3PO2}$ and not $\ce{LiAlH4}$ ?
Also, what is the reaction mechanism of the reaction with $\ce{H3PO2}$
?

This Wikipedia page list all kind of nucleophilic substitution reaction on aromatic compound why can't $\ce{LiAlH4}$ / $\ce{H-}$ react like one of these to give benzene and expel nitrogen.
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.

Update
I was just researching (on google) for the mechanism where I came across this site suggesting the mechanism to be radical and as follow :

But still there is one problem the side products don't match this site states the side product to be $\ce{H3PO3}$ and $\ce{HCl}$ but as stated above the side products must be $\ce{H2PXO2}$.
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.