$7.60 \log _{2}(2-x)-\log _{2}(2-\sqrt{x})=\log _{2} \sqrt{2-x}-0.5$.
7.60 Given the domain of the logarithmic function, we have

$$
\left\{\begin{array} { l } 
{ 2 - x > 0 } \\
{ 2 - \sqrt { x } > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x < 2 \\
x < 4
\end{array}\right.
\right.
$$

From the above, we get \(0 < x < 2\) (since \(x < 2\) is the stricter condition). Given the equation \(\frac{\sqrt{2-x}}{2-\sqrt{x}}=\frac{1}{\sqrt{2}}\), we solve it by squaring both sides and substituting \(\sqrt{x}=y \geq 0\). As a result, we find its roots \(x_{1}=0\), \(x_{2}=\frac{16}{9}\), and both satisfy condition (1).

Answer: \(x_{1}=0, x_{2}=\frac{16}{9}\).