Three, (15 points) Given real numbers $x, y$ satisfy the conditions:
$$
\left\{\begin{array}{l}
x+2 y>0, \\
x-2 y>0, \\
(x+2 y)(x-2 y)=4 .
\end{array}\right.
$$

Find the minimum value of $|x|-|y|$.
From the problem, we have
\[
\left\{
\begin{array}{l}
x > 2|y| \geqslant 0, \\
x^{2} - 4 y^{2} = 4.
\end{array}
\right.
\]
It is clear that \( x > 0 \), and by symmetry, we can assume \( y \geqslant 0 \).
Thus, \( |x| - |y| = x - y \).
Let \( u = x - y \). Then, from \( x > 2 y \geqslant 0 \), we know \( u > 0 \).
Substituting \( x = u + y \) into \( x^{2} - 4 y^{2} = 4 \), we get
\[
3 y^{2} - 2 u y + 4 - u^{2} = 0.
\]

Since \( y \in \mathbf{R} \), we have
\[
\Delta = 4 u^{2} - 12(4 - u^{2}) \geqslant 0 \Rightarrow u \geqslant \sqrt{3}.
\]

When \( u = \sqrt{3} \), the solution to equation (1) is \( y = \frac{\sqrt{3}}{3} (>0) \).
Thus, \( x = \frac{4 \sqrt{3}}{3} \).
In summary, the minimum value of \( |x| - |y| \) is \( \sqrt{3} \).