解:
(1)
$\begin{aligned}&(2x - 3)m+2m^{2}-3x\\=&2mx-3m + 2m^{2}-3x\\=&(2m - 3)x+2m^{2}-3m\end{aligned}$
因为其值与$x$的取值无关,所以$2m - 3 = 0,$
解得$m=\frac{3}{2}。$
(2)
因为$A=(2x + 1)(x - 1)-x(1 - 3y),$$B=-x^{2}+xy - 1,$
所以
$\begin{aligned}&3A + 6B\\=&3[(2x + 1)(x - 1)-x(1 - 3y)]+6(-x^{2}+xy - 1)\\=&3(2x^{2}-2x+x - 1-x + 3xy)-6x^{2}+6xy - 6\\=&6x^{2}-6x + 3x-3-3x + 9xy-6x^{2}+6xy - 6\\=&15xy-6x - 9\\=&3x(5y - 2)-9\end{aligned}$
因为$3A + 6B$的值与$x$的取值无关,
所以$5y - 2 = 0,$即$y=\frac{2}{5}。$
(3)设$AB = x,$由题图可知$S_{1}=a(x - 3b),$$S_{2}=2b(x - 2a),$
所以
$\begin{aligned}S_{1}-S_{2}&=a(x - 3b)-2b(x - 2a)\\&=ax-3ab-2bx + 4ab\\&=(a - 2b)x+ab\end{aligned}$
因为当$AB$的长变化时,$S_{1}-S_{2}$的值始终保持不变,
所以$S_{1}-S_{2}$的值与$x$无关,
所以$a - 2b = 0,$所以$a = 2b。$