$8.52 \frac{4}{\sqrt{2 - x}}-\sqrt{2-x}<2$.
8.52 Assuming $y=\sqrt{2-x}$, we obtain the inequality

$\frac{4}{y}-y>0$ and, therefore, we can multiply both sides of inequality (1) by $y$. Consequently,

$4-y^{2}>0$.

From this, we find $y<\sqrt{5}-1$. However, $y>0$. Thus, $y>\sqrt{5}-1$ and it remains to solve the inequality $\sqrt{2-x}>\sqrt{5}-1$. Since $\sqrt{5}-1>0$, by reference to $6^{0}$, this irrational inequality is equivalent to the system

$$
\left\{\begin{array} { l } 
{ 2 - x > 0 } \\
{ ( \sqrt { 2 - x } ) ^ { 2 } > ( \sqrt { 5 } - 1 ) ^ { 2 } }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x < 2 } \\
{ x < 6 - 2 \sqrt { 5 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x<2 \\
x<2 \sqrt{5}-4
\end{array}\right.\right.\right.
$$

from which we obtain that $x<2 \sqrt{5}-4$.

Solution: $(-\infty, 2 \sqrt{5}-4)$.