$\left(1\right)\because x^{2}-4x-5=0$，
因式分解得$\left(x+1\right)\left(x-5\right)=0$，
$\therefore x+1=0$或$x-5=0$，
$\therefore x_{1}=-1$，$x_{2}=5$；
$(2)\because \left(4y-1\right)^{2}-25=0$，
$\therefore \left(4y-1\right)^{2}=25$，
$\therefore$开方得$4y-1=\pm 5$，
$\therefore y_{1}=-1$，${y}_{2}=\frac{3}{2}$；
$(3)\because 3x^{2}=4x-1$，
$\therefore 3x^{2}-4x+1=0$，
$\therefore$因式分解得$\left(x-1\right)\left(3x-1\right)=0$，
$\therefore x-1=0$或$3x-1=0$，
$\therefore x_{1}=1$，${x}_{2}=\frac{1}{3}$；
$(4)\because \left(x-3\right)^{2}+2x\left(x-3\right)=0$，
$\therefore$因式分解得$\left(x-3\right)\left(x-3+2x\right)=0$，
$\therefore x-3=0$或$3x-3=0$，
$\therefore x_{1}=3$，$x_{2}=1$.

$\left(1\right)\because x^{2}-4x-5=0$，
因式分解得$\left(x+1\right)\left(x-5\right)=0$，
$\therefore x+1=0$或$x-5=0$，
$\therefore x_{1}=-1$，$x_{2}=5$；
$(2)\because \left(4y-1\right)^{2}-25=0$，
$\therefore \left(4y-1\right)^{2}=25$，
$\therefore$开方得$4y-1=\pm 5$，
$\therefore y_{1}=-1$，${y}_{2}=\frac{3}{2}$；
$(3)\because 3x^{2}=4x-1$，
$\therefore 3x^{2}-4x+1=0$，
$\therefore$因式分解得$\left(x-1\right)\left(3x-1\right)=0$，
$\therefore x-1=0$或$3x-1=0$，
$\therefore x_{1}=1$，${x}_{2}=\frac{1}{3}$；
$(4)\because \left(x-3\right)^{2}+2x\left(x-3\right)=0$，
$\therefore$因式分解得$\left(x-3\right)\left(x-3+2x\right)=0$，
$\therefore x-3=0$或$3x-3=0$，
$\therefore x_{1}=3$，$x_{2}=1$.