pH of 0.05 M NaH2PO3
The problem with method 1 is that you discard the protonation process, which effectively would increase pH as it produces $\ce{OH-}$ ions.

The criteria of $K_1 &gt; 1000K_2$ is applicable when you have a polyprotic acid in its completely (de)protonated state. The ionisation of H2A produces only a small amount of HA- and K2 is small as well, so ionisation of HA- would be negligible and we can ignore it.

However with an ampholytic species, while K2 is smaller than K1, the initial reactant is the same (HA-) and it is in high concentration, so both processes occur. The formula in method 2 is the result of this consideration.