解:【理解应用】
(1)因为代数式$(2m - 3)x+2m^{2}-3m$的值与$x$的取值无关,所以$2m - 3 = 0,$解得$m=\frac{3}{2}。$
(2)
$\begin{aligned}&3[(2x + 1)(x - 1)-x(1 - 3y)]+6(-x^{2}+xy - 1)\\=&3(2x^{2}-2x+x - 1-x + 3xy)+(-6x^{2}+6xy - 6)\\=&3(2x^{2}-2x + 3xy - 1)-6x^{2}+6xy - 6\\=&6x^{2}-6x + 9xy - 3-6x^{2}+6xy - 6\\=&(-6x)+(9xy + 6xy)+(6x^{2}-6x^{2})+(-3 - 6)\\=&(-6x)+15xy-9\end{aligned}$
因为代数式的值与$x$的取值无关,所以$-6 + 15y = 0,$解得$y=\frac{2}{5}。$
【能力提升】
设$AB$的长为$x,$$S_{1}=b(x - 3b),$$S_{2}=a(x - 2a),$
$S_{1}-S_{2}=b(x - 3b)-a(x - 2a)=bx-3b^{2}-ax + 2a^{2}=(b - a)x+2a^{2}-3b^{2},$
因为$S_{1}-S_{2}$的值始终保持不变,所以$b - a = 0,$即$a = b。$