（1）把点$P\left(2,n\right)$代入$y_{2}=-4x+12$得：$n=-8+12=4$，
$\therefore P\left(2,4\right)$，
把$A\left(0,2\right)$，$P\left(2,4\right)$代入$y_{1}=kx+b$得，$\left\{{\begin{array}{l}{b=2}\\{2k+b=4}\end{array}}\right.$，
解得：$\left\{{\begin{array}{l}{k=1}\\{b=2}\end{array}}\right.$，
$\therefore y_{1}=x+2$，
把$B\left(m,0\right)$代入$y_{1}=x+2$得：$0=m+2$，
解得：$m=-2$，
$\therefore m=-2$，$n=4$；
$(2)\because$直线$y_{1}=kx+b$与$y_{2}=-4x+12$交于点$P\left(2,4\right)$，
$\therefore$方程组$\left\{{\begin{array}{l}{y=kx+b}\\{y=-4x+12}\end{array}}\right.$的解为：$\left\{{\begin{array}{l}{x=2}\\{y=4}\end{array}}\right.$，
故答案为：$\left\{{\begin{array}{l}{x=2}\\{y=4}\end{array}}\right.$；
$(3)$当$y_{2}=-4x+12=0$时，
解得：$x=3$，
$\therefore C\left(3,0\right)$，
$\because P\left(2,4\right)$，$B\left(-2,0\right)$，$C\left(3,0\right)$，
$\therefore BC=5$，
$\therefore {S_{△PBC}}=\frac{1}{2}×5×4=10$.

（1）把点$P\left(2,n\right)$代入$y_{2}=-4x+12$得：$n=-8+12=4$，
$\therefore P\left(2,4\right)$，
把$A\left(0,2\right)$，$P\left(2,4\right)$代入$y_{1}=kx+b$得，$\left\{{\begin{array}{l}{b=2}\\{2k+b=4}\end{array}}\right.$，
解得：$\left\{{\begin{array}{l}{k=1}\\{b=2}\end{array}}\right.$，
$\therefore y_{1}=x+2$，
把$B\left(m,0\right)$代入$y_{1}=x+2$得：$0=m+2$，
解得：$m=-2$，
$\therefore m=-2$，$n=4$；
$(2)\because$直线$y_{1}=kx+b$与$y_{2}=-4x+12$交于点$P\left(2,4\right)$，
$\therefore$方程组$\left\{{\begin{array}{l}{y=kx+b}\\{y=-4x+12}\end{array}}\right.$的解为：$\left\{{\begin{array}{l}{x=2}\\{y=4}\end{array}}\right.$，
故答案为：$\left\{{\begin{array}{l}{x=2}\\{y=4}\end{array}}\right.$；
$(3)$当$y_{2}=-4x+12=0$时，
解得：$x=3$，
$\therefore C\left(3,0\right)$，
$\because P\left(2,4\right)$，$B\left(-2,0\right)$，$C\left(3,0\right)$，
$\therefore BC=5$，
$\therefore {S_{△PBC}}=\frac{1}{2}×5×4=10$.