阅读探索：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$\left\{\begin{array}{l}a-1=2,\\ b+2=2.\\\end{array}\right.\therefore\left\{\begin{array}{l}a=3,\\ b=0.\\\end{array}\right.$

此种解方程组的方法叫换元法.

$(1)$拓展提高：

运用上述方法解方程组$\left\{\begin{array}{l}(\dfrac{a}{3}-1)+2(\dfrac{b}{5}+2)=4,\\ 2(\dfrac{a}{3}-1)+(\dfrac{b}{5}+2)=5.\\\end{array}\right.$

$(2)$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$\left\{\begin{array}{l}x=5,\\ y=3,\\\end{array}\right.$

求关于$m$，$n$的方程组$\left\{\begin{array}{l}5{{a}_{1}}(m+3)+3{{b}_{1}}(n-2)={{c}_{1}},\\ 5{{a}_{2}}(m+3)+3{{b}_{2}}(n-2)={{c}_{2}}\\\end{array}\right.$的解.