$7.1 \quad \sqrt{5} \cdot 0.2^{\frac{1}{2 x}}-0.04^{1-x}=0$.
7.1 Here all powers can be reduced to the same base 5. We have:

$$
\begin{aligned}
& \sqrt{5}=5^{\frac{1}{2}}, 0.2^{\frac{1}{2 x}}=\left(\frac{1}{5}\right)^{\frac{1}{2 x}}=5^{-\frac{1}{2 x}} \\
& 0.04^{1-x}=\left(\frac{1}{25}\right)^{1-x}=5^{-2(1-x)}
\end{aligned}
$$

Then the equation will take the form

$$
5^{\frac{1}{2}} \cdot 5^{-\frac{1}{2 x}}-5^{-2(1-x)}=0, \text { or } 5^{\frac{1}{2}-\frac{1}{2 x}}=5^{-2(1-x)}
$$

According to the hint $1^{0}$, we transition to the equivalent equation

$$
\frac{1}{2}-\frac{1}{2 x}=-2(1-x)
$$

After transformations, we get

$\left\{\begin{array}{l}4 x^{2}-5 x+1=0, \\ x \neq 0,\end{array}\right.$

from which $x_{1}=1, x_{2}=\frac{1}{4}$.

Answer: $x_{1}=1, x_{2}=0.25$.