(1)∵|a-2|+(b-3)2=0, ∴a-2=0,b-3=0, 即a=2,b=3, 又∵c=2b-a, ∴c=2×3-2=4;
(2)由题意:S△ABC=BC×b=×4×3=6, S四边形ABOP=×AO×|m|+×AO×|c|=×2×|m|+×2×3=|m|+3, 由题意S四边形ABOP=S△ABC, ∴|m|+3=6, 即m=±3, ∵点P在第二象限, ∴点P(-3,1);
(3)∠AQB为定值. 证明:∵2∠BAQ=∠AOB+∠ABO,2∠ABQ=∠AOB+∠OAB, ∴2(∠BAQ+∠ABQ)=2∠AOB+∠ABO+∠OAB, ∠BAQ+∠ABQ=∠AOB+=90°+∠AOB, ∵∠AOB大小为定值, ∴∠BAQ+∠ABQ的大小为定值, ∴∠AQB=180°-(∠BAQ+∠ABQ), 故∠AQB为定值. |