(2) Xiao Cong's approach is correct.

The reasons are as follows: in Rt△OMP and Rt△ONP, op = opom = on.

∴Rt△OMP≌Rt△ONP(HL)，

∴∠MOP=∠NOP，

∴OP is the bisector of ∞∠AOB;

(3)① As shown in Figure 2, draw points M and N on OA and OB with the scale on the scale, so that OM = ON.

② Make MP=NP with two scales, and cross at point P;

③ As a ray OP, OP is the bisector of ∠AOB.

The reasons are as follows: in △MOP and △NOP, OM = 0 NOP = OPMP = NP.

∴△MOP≌△NOP(SSS)，

∴∠MOP=∠NOP，

∴OP is the bisector of ∞∠AOB.