,- 1.3 .
Remove the logarithm, add the three equations to get
$$
3 x y z+\log _{5} x y z=378 \text {. }
$$

Considering $f(x)=3 x+\log _{5} x$ is strictly monotonically increasing, and $f(125)=378$, then
$$
x y z=125 \Rightarrow \log _{5} x y z=3 .
$$
1. Let $x, y, z$ satisfy:
$$
\left\{\begin{array}{l}
\log _{2}\left(x y z-3+\log _{5} x\right)=5, \\
\log _{3}\left(x y z-3+\log _{5} y\right)=4, \\
\log _{4}\left(x y z-3+\log _{5} z\right)=4 .
\end{array}\right.
$$

Then $\log _{5} x y z=$ $\qquad$