阅读探索.
知识累积:解方程组$\left\{\begin{array}{l}(a-1)+2(b+2)=6,\\ 2(a-1)+(b+2)=6,\end{array}\right.$,
解:设$a-1=x$,$b+2=y$,原方程组可变为$\left\{\begin{array}{l}x+2y=6\\ 2x+y=6\end{array}\right.$
解方程组,得:$\left\{\begin{array}{l}x=2\\ y=2\end{array}\right.$,即$\left\{\begin{array}{l}a-1=2\\ b+2=2\end{array}\right.$,解得$\left\{\begin{array}{l}a=3\\ b=0\end{array}\right.$.此种解方程组的方法叫换元法.
$(1)$举一反三:运用上述方法解下列方程组:$\left\{\begin{array}{l}(\frac{a}{3}-1)+2(\frac{b}{5}+2)=4\\ 2(\frac{a}{3}-1)+(\frac{b}{5}+2)=5\end{array}\right.$；
$(2)$能力运用:已知关于$x$,$y$的方程组$\left\{\begin{array}{l}{a}_{1}x+{b}_{1}y={c}_{1}\\{a}_{2}x+{b}_{2}y={c}_{2}\end{array}\right.$的解为$\left\{\begin{array}{l}x=5\\ y=3\end{array}\right.$,则关于$m$,$n$的方程组$\left\{\begin{array}{l}{a}_{1}(m+3)+{b}_{1}(n-2)={c}_{1}\\{a}_{2}(m+3)+{b}_{2}(n-2)={c}_{2}\end{array}\right.$的解是______；
$(3)$拓展提高:若方程组$\left\{\begin{array}{l}3{a}_{1}x+2{b}_{1}y=5{c}_{1}\\ 3{a}_{2}x+2{b}_{2}y=5{c}_{2}\end{array}\right.$的解是$\left\{\begin{array}{l}x=3\\ y=4\end{array}\right.$,则方程组$\left\{\begin{array}{l}{a}_{1}x+{b}_{1}y={c}_{1}\\{a}_{2}x+{b}_{2}y={c}_{2}\end{array}\right.$的解是______.