1. If $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $|x|-|y|$ is $\qquad$
$=\sqrt{1} \cdot \sqrt{3}$.
From the problem, we have
$$
\left\{\begin{array}{l}
x + 2 y > 0, \\
x - 2 y > 0, \\
(x + 2 y)(x - 2 y) = 4.
\end{array} \text { Then } \left\{\begin{array}{l}
x > 2|y| \geqslant 0, \\
x^{2} - 4 y^{2} = 4.
\end{array}\right.\right.
$$

By its symmetry, we only need to consider $y \geqslant 0$.
Since $x > 0$, we only need to find the minimum value of $x - y$.
Let $x - y = u$, substitute into $x^{2} - 4 y^{2} = 4$, we get
$$
3 y^{2} - 2 u y + (4 - u^{2}) = 0.
$$

The above equation is a quadratic equation in $y$, and it clearly has real roots. Therefore, $\Delta = 16(u^{2} - 3) \geqslant 0$. Solving this, we get $u \geqslant \sqrt{3}$.
When $x = \frac{4 \sqrt{3}}{3}, y = \frac{\sqrt{3}}{3}$, $u = \sqrt{3}$.