我们约定:不等式组$m \lt x \lt n$,$m \lt x\leqslant n$,$m\leqslant x \lt n$,$m\leqslant x\leqslant n$的"长度"均为$d=n-m$,$\left(m \lt n\right)$,不等式组的整数解称为不等式组的"整点".例如:$-2 \lt x\leqslant 2$的"长度"$d=2-\left(-2\right)=4$,"整点"为$x=-1$,$0$,$1$,$2$.根据该约定,解答下列问题:
$(1)$不等式组$\left\{\begin{array}{l}5x+3＞3x\\ 2x-1≤0\end{array}\right.$的"长度"$d=$______；"整点"为______；
$(2)$若不等式组$\left\{\begin{array}{l}1≤x≤3\\ ax-3＜\frac{1}{2}x+1\end{array}\right.$的"长度"$d=2$,求$a$的取值范围；
$(3)$若不等式组$\left\{\begin{array}{l}1≤x≤3\\ a≤x≤\frac{1}{2}a+2\end{array}\right.$的"长度$d=\frac{3}{2}$,此时是否存在实数$m$使得关于$y$的不等式组$\left\{\begin{array}{l}y+1＞m\\ ay-1≤2m\end{array}\right.$恰有$4$个"整点",若存在,求出$m$的取值范围；若不存在,请说明理由.