(1) vector OA+2 vector OB+3 vector OC=0 vector, and then extend OB and OC to E and F respectively, so that o E = 2OB, OF=3OC, then the vector sum OF OA, OE and OF is 0, so that OA, OE and of can be translated to form a triangle, and OA: OE: of can be easily known.

So sinboc: sinoac: sinaob = OA: 20b: 3oc.

So S△AOB:S△BOC:S△COA.

=(OA * OB * sinAOB):(OB * OC * sinBOC):(OA * OC * sinAOC)=(OA * OB * 3OC):(OB * OC * OA):(OA * OC * 2OB)= 3: 1:2

(2) Similarly, by using Lamy theorem, there are SINAOB: SINBOC: SINCOA = (n+m) OC: NOA: MOB.

So S△AOB:S△BOC:S△COA.

=(OA * OB * sinAOB):(OB * OC * sinBOC):(OA * OC * sin coa)=(OA * OB *(n+m)OC):(OB * OC * nOA):(OA * OC * mOB)=(n+m):n:m