（1）根据两点的距离公式得，$|{PQ}|=\sqrt{{{({2+3})}^2}+{{({6+6})}^2}}=\sqrt{25+144}=13$；

$(2)\left(m-1\right)^{2}+9=25$，

$\therefore m_{1}=5$，$m_{2}=-3$；

$(3)\because \sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$看成点$\left(x,y\right)$到两点$\left(3,0\right)$和$\left(-3,-4\right)$的距离之和，

$\therefore \sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$的最小值为点$\left(x,y\right)$到两点$\left(3,0\right)$和$\left(-3,-4\right)$的距离之和的最小值，

$\because$当点$\left(x,y\right)$在以两点$\left(3,0\right)$和$\left(-3,-4\right)$为端点的线段上时，点$\left(x,y\right)$到两点$\left(3,0\right)$和$\left(-3,-4\right)$的距离之和的最小值，其最小值为以两点$\left(3,0\right)$和$\left(-3,-4\right)$为端点的线段长度，

$\therefore \sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$的最小值为$\sqrt{{{({3+3})}^2}+{{({0+4})}^2}}=2\sqrt{13}$.

（1）根据两点的距离公式得，$|{PQ}|=\sqrt{{{({2+3})}^2}+{{({6+6})}^2}}=\sqrt{25+144}=13$；

$(2)\left(m-1\right)^{2}+9=25$，

$\therefore m_{1}=5$，$m_{2}=-3$；

$(3)\because \sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$看成点$\left(x,y\right)$到两点$\left(3,0\right)$和$\left(-3,-4\right)$的距离之和，

$\therefore \sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$的最小值为点$\left(x,y\right)$到两点$\left(3,0\right)$和$\left(-3,-4\right)$的距离之和的最小值，

$\because$当点$\left(x,y\right)$在以两点$\left(3,0\right)$和$\left(-3,-4\right)$为端点的线段上时，点$\left(x,y\right)$到两点$\left(3,0\right)$和$\left(-3,-4\right)$的距离之和的最小值，其最小值为以两点$\left(3,0\right)$和$\left(-3,-4\right)$为端点的线段长度，

$\therefore \sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$的最小值为$\sqrt{{{({3+3})}^2}+{{({0+4})}^2}}=2\sqrt{13}$.