5. $\frac{5}{3}$.

Obviously,
$$
\begin{array}{c}
a z_{0}+2 b y_{0}+c x_{0}=0,4 z_{0}^{2}+x_{0}^{2}=4 y_{0}^{2} \\
\Rightarrow 4 b^{2} y_{0}^{2}=\left(a z_{0}+c x_{0}\right)^{2} .
\end{array}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(x_{0}^{2}+4 z_{0}\right)\left(c^{2}+\frac{a^{2}}{4}\right) \geqslant\left(c x_{0}+a z_{0}\right)^{2} \\
\Rightarrow 4 y_{0}^{2}\left(c^{2}+\frac{a^{2}}{4}\right) \geqslant 4 b^{2} y_{0}^{2} \\
\Rightarrow c^{2}+\frac{a^{2}}{4} \geqslant b^{2} . \\
\text { and } a^{2}+c^{2}>c^{2}+\frac{a^{2}}{4} \geqslant b^{2}(b>a, b>c),
\end{array}
$$

and $a^{2}+c^{2}>c^{2}+\frac{a^{2}}{4} \geqslant b^{2}(b>a, b>c)$,
thus, $\triangle A B C$ is an acute triangle.
By the Law of Sines, we have
$$
\begin{array}{l} 
\sin ^{2} C+\frac{1}{4} \sin ^{2} A \geqslant \sin ^{2} B \\
\Rightarrow \frac{1}{4} \sin ^{2} A \geqslant \sin ^{2} B-\sin ^{2} C \\
=\sin (B+C) \cdot \sin (B-C) \\
=\sin A \cdot \sin (B-C) \\
\Rightarrow \sin A \geqslant 4 \sin (B-C) \\
\Rightarrow \sin B \cdot \cos C+\sin C \cdot \cos B \\
\quad \geqslant 4 \sin B \cdot \cos C-4 \sin C \cdot \cos B \\
\Rightarrow 5 \sin C \cdot \cos B \geqslant 3 \sin B \cdot \cos C \\
\Rightarrow \tan B \cdot \cot C \leqslant \frac{5}{3} .
\end{array}
$$
5. Given that the three sides of $\triangle A B C$ are $a, b, c$ respectively, and $b>\max \{a, c\}$, there are three non-zero real numbers $x_{0}, y_{0}, z_{0}$ satisfying that the line $a x+b y+c=0$ passes through the point $\left(\frac{z_{0}}{x_{0}}, \frac{2 y_{0}}{x_{0}}\right)$. If the point $\left(\frac{z_{0}}{y_{0}}, \frac{x_{0}}{y_{0}}\right)$ lies on the ellipse $x^{2}+\frac{y^{2}}{4}=1$, then the maximum value of $\tan B \cdot \cot C$ is . $\qquad$