balance the 2nd one using oxidation no method.

Dear Student,
To balance by oxidation number method first identify the species undergoing the change in oxidation state.
P+OH^-$\to$PH₃+H₂PO₂
Here P has an oxidation state of 0 on the reactant side it is undergoing change to -3 as in PH₃
P=0 is also changing to +2 state as in H₂PO2
$$\begin{gathered} P^0\to P^{-3}{H}_3 (reduction ) change of 3 \\ {P}^0\to H_2{P}^{+2}{O}_2 (oxidation ) change of 2 \end{gathered}$$
Balance the charges on both sides by adding equal number of OH^- on both sides
P +3H₂O$\to$PH₃+3OH^-
P+2OH^​-$\to$H₂PO₂
Multiply equation 1 with 2 and equation 2 with 3. and cancel both OH^- on both sides as the charges are equal
Therefore overall balanced equation will be 

$5P+6H_{2}O\to 2P{H}_3+3H_{2}P{O}_2$
Regards!