The oxidising agent and reducing agent in the given reaction are

$5{\mathrm{P}}_{4\left(\mathrm{s}\right)}+12{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+12{{\mathrm{HO}}^{-}}_{\left(\mathrm{aq}\right)}\to$
$8{\mathrm{PH}}_{3\left(\mathrm{g}\right)}+12{{\mathrm{HPO}}_2^{-}}_{\left(\mathrm{aq}\right)}$

1. Oxidising agent = ${\mathrm{P}}_4$; Reducing agent = ${\mathrm{P}}_4$

2. Oxidising agent = ${\mathrm{P}}_4$; Reducing agent = ${\mathrm{H}}_2\mathrm{O}$

3. Oxidising agent = ${\mathrm{H}}_2\mathrm{O}$; Reducing agent = ${\mathrm{P}}_4$

4. None of the above

The set of metals that can show disproportionation reaction is:
1. Cu, Na, Li
2. Mg, F, Ne
3. P, Cl, S
4. Mn, Cu, Ga