1. Multiplying the first and fourth, and the second and third factors, we get the equation: $\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)=360$.

By making the substitution $y=x^{2}+5 x+4$, we get $y^{2}+2 y-360=0$, from which $y_{1}=-20$, $y_{2}=18$. Therefore, we have the equations:

$x^{2}+5 x+24=0, x^{2}+5 x-14=0$

The first equation has no solutions, from the second we get: $x_{1}=-7, x_{2}=2$.
Answer: $x_{1}=-7, x_{2}=2$