阅读理解:
定义:若一元一次方程的解在一元一次不等式组解集范围内,则称该一元一次方程为该不等式组的"子方程".例如:$2x-1=3$的解为$x=2$,$\left\{\begin{array}{l}2x-3＜9-x\\ 5x+5≥2x-4\end{array}\right.$的解集为$-3\leqslant x \lt 4$,不难发现$x=2$在$-3\leqslant x \lt 4$的范围内,所以$2x-1=3$是$\left\{\begin{array}{l}2x-3＜9-x\\ 5x+5≥2x-4\end{array}\right.$的"子方程".
问题解决:
$(1)$在方程①$3x-1=0$,②$\frac{2}{3}x-1=0$,③$2x+3\left(x+2\right)=21$中,不等式组$\left\{\begin{array}{l}2x-1＞x+1\\ 3(x-2)-x≤4\end{array}\right.$的"子方程"是______$;(填序号)$
$(2)$若关于$x$的方程$2x-k=2$是不等式组$\left\{\begin{array}{l}3x-6＞4-x\\ x-1≥4x-10\end{array}\right.$的"子方程",求$k$的取值范围；
$(3)$若方程$2x+4=0$,$\frac{2x-1}{3}=-1$都是关于$x$的不等式组$\left\{\begin{array}{l}(m-2)x＜m-2\\ x+5≥m\end{array}\right.$的"子方程",直接写出$m$的取值范围.