定义:若一元一次方程的解在一元一次不等式组的解集范围内,则称该一元一次方程为该不等式组的"子方程".例如:$2x+4=2$的解为$x=-1$,$\left\{\begin{array}{l}{2x-3＜9-x}\\{5x+5≥2x-4}\end{array}\right.$集为$-3\leqslant x \lt 4$,不难发现$x=-1$在$-3\leqslant x \lt 4$的范围内,所以$2x+4=2$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$4x-5=x+7$,②$\frac{1}{11}x-\frac{1}{3}=0$,③$2x+3\left(x+2\right)=21$中,不等式组$\left\{\begin{array}{l}{2x-1＞-x+8}\\{3(x-2)-x≤4}\end{array}\right.$的"子方程"是______$(填序号)$；
$(2)$若方程$4x+4=0$是关于$x$的不等式组$\left\{\begin{array}{l}{2x+8≥m}\\{3x+m＜2m+9}\end{array}\right.$的"子方程",试求$m$的取值范围；
$(3)$若关于$x$的方程$2x-k=4$接不等式组$\left\{\begin{array}{l}{5x-7＞11-x}\\{\frac{1}{3}x≥\frac{1}{2}x-1}\end{array}\right.$的"子方程",求$E$的取值范围.