$\because a$与$b$互为相反数$\left(a\neq 0\right)$，
$\therefore \frac{b}{a}=-1$，
$\because c$与$d$互为倒数，
$\therefore cd=1$，
$\because m$的绝对值是$2$，
$\therefore m=\pm 2$，
$\therefore m^{2}=4$，
$\because \left(x-2\right)^{2}+|y+2|=0$，
又$\because \left(x-2\right)^{2}\geqslant 0$，$|y+2|\geqslant 0$，
$\therefore x-2=0$，$y+2=0$，
$\therefore x=2$，$y=-2$，
$\therefore x-y=2-\left(-2\right)=2+2=4$，
$\therefore (x-y)^{2}+(\frac{b}{a})^{2025}-(-cd)^{4}+m^{2}$
$=4^{2}+\left(-1\right)^{2025}-\left(-1\right)^{4}+4$
$=16-1-1+4$
$=18$.

$\because a$与$b$互为相反数$\left(a\neq 0\right)$，
$\therefore \frac{b}{a}=-1$，
$\because c$与$d$互为倒数，
$\therefore cd=1$，
$\because m$的绝对值是$2$，
$\therefore m=\pm 2$，
$\therefore m^{2}=4$，
$\because \left(x-2\right)^{2}+|y+2|=0$，
又$\because \left(x-2\right)^{2}\geqslant 0$，$|y+2|\geqslant 0$，
$\therefore x-2=0$，$y+2=0$，
$\therefore x=2$，$y=-2$，
$\therefore x-y=2-\left(-2\right)=2+2=4$，
$\therefore (x-y)^{2}+(\frac{b}{a})^{2025}-(-cd)^{4}+m^{2}$
$=4^{2}+\left(-1\right)^{2025}-\left(-1\right)^{4}+4$
$=16-1-1+4$
$=18$.