解:
$ (1)(-2ma^2 + 3a -1) + [-4a^2 + (n-1)a -1]$
$ =-2ma^2 + 3a -1 -4a^2 + (n-1)a -1$
$ =(-2m -4)a^2 + (3 + n -1)a -2$
由题意,得$-2m -4 = 0$且$3 + n -1 = 0$,
解得$m=-2$,$n=-2$。
$ (2) $因为$(4\ \mathrm {m^2}n - 3mn^2) - 2(\mathrm {m^2}n + mn^2)$
$ =4\ \mathrm {m^2}n - 3mn^2 - 2\ \mathrm {m^2}n - 2mn^2$
$ =2\ \mathrm {m^2}n -5mn^2$
$ $由$(1)$知$m=-2$,$n=-2$,
$ $所以原式$=2×(-2)^2×(-2) -5×(-2)×(-2)^2$
$ =-16 + 40 = 24$