∵ Only OP=OP and PM=PN cannot judge △ OPM △ OPN;
∴ It cannot be judged that OP is the bisector of ∞∠AOB;
(2) Scheme (2) is feasible,
Proof: at △OPM and △OPN
∵OM=ONOP=OPPM=PN，
∴△OPM≌△OPN(SSS)，
∴∠ AOP =∠∠ BOP (the angles corresponding to congruent triangles are equal),
∴OP is the bisector of ∞∠AOB.
(3) Solution: ① If OC is inside ∠AOB (Figure 1),
∫∠AOC:∠COB = 1:3，
∴ let ∠AOC=x, ∠COB=3x,
∫∠AOB = 60，
∴x+3x=60，
X = 15，
That is ∠ AOC = 15,
∫OP shares ∠AOB,
∴∠AOP=30，
∴∠cop=∠aod-∠aoc=30- 15 = 15。
② If OC is outside ∠AOB (as shown in Figure 2),
∫∠AOC:∠COB = 1:3，
∴ let ∠AOC=x, ∠COB=3x,
∫∠AOB = 60，
∴3x-x=60，
X = 30，
∴∠AOC=x=30，∠COB=3x=3×30 =90，
∫OP shares ∠AOB,
∴∠AOP=30，
∴∠COP=∠AOC+∠AOP=30 +30 =60。
The angle formed by the bisector of ∴OC and ∠AOB is15 or 60.