∴ angle ABM= angle OMB

∵BM bisector ABC

∴ angle ABM= angle CBM

OM parallel to BC

∴ angle AMO= angle AEB

In triangle ABC, AB=AC, AE is the angular bisector.

Ae is perpendicular to BC.

∴ Angle aeb = 90

∴ Angle δ ω = 90.

∴OM perpendicular to AE

∴AE is tangent to circle O.

2. In the triangle ABC, AB=AC, AE is the bisector of the angle.

∴BE= 1/2BC, angle ABC= angle C.

BC = 4，cosC= 1/3

∴BE=2, cos angle ABC= 1/3.

In △ABC, the angle ABE = 90.

So AB=BE/cos angle ABC=6.

Let the radius of circle o be r, then ao = 6-r.

OM parallel to BC

∴△AOM∽△ABE

∴OM/BE=AO/AB

∴r/2=(6-r)/6

The solution is r=3/2.

∴: The radius of circle O is 3/2.

I tried to type with symbols, but some symbols couldn't be typed. You can make do with it. . . This is hard work. . .