(1)解:因?yàn)?A = 2a^{2}+3ab - 2a - 1$,$B=-a^{2}+ab + 1$,所以$A + 2B=(2a^{2}+3ab - 2a - 1)+2(-a^{2}+ab + 1)$
$=2a^{2}+3ab - 2a - 1 - 2a^{2}+2ab + 2$
$=(2a^{2}-2a^{2})+(3ab + 2ab)-2a+(-1 + 2)$
$=5ab - 2a + 1$
當(dāng)$a=-1$,$b = 2$時(shí),原式$=5×(-1)×2-2×(-1)+1=-10 + 2 + 1=-7$
(2)解:由(1)知$A + 2B=(5b - 2)a+1$,因?yàn)榇鷶?shù)式的值與$a$的取值無(wú)關(guān),所以$5b-2 = 0$,解得$b=\frac{2}{5}$