解:?$(1)$?原式?$=4a2+a-5-3a-3a2=a2-2a-5$?
∵?$a2-2a-1=0$?,∴?$a2-2a=1$?,原式?$=1-5=-4$?
?$(2)$?∵?$(a-2)2$?和?$|b+\frac {1}{2}|$?的值都是非負數(shù)
∴?$a-2=0$?,?$b+\frac {1}{2}=0$?
∴?$a=2$?,?$b=-\frac {1}{2}$?
?$ $?原式?$=3a^2b - (2a2-ab2+3a2b-4ab2) =3a2b-2a2+ab2-3a2b+4ab2 =5ab2-2a2$?
將?$a=2$?,?$b=-\frac {1}{2}$?代入得,原式?$=5×2×\frac {1}{4}-2×4=2.5-8=-5.5$?