1. The range of real numbers $x$ that satisfy $\sqrt{1+x}+\frac{2}{5}<\sqrt{3-x}$ is $\qquad$
$-1 \cdot\left[-1, \frac{11}{25}\right)$.

From the problem, we know $-1 \leqslant x \leqslant 3$, and
$$
\sqrt{3-x}-\sqrt{1+x}>\frac{2}{5} \text {. }
$$

Combining $(3-x)-(1+x)=2-2 x$, we get
$$
\sqrt{3-x}+\sqrt{1+x}0 \\
\Rightarrow(25 x-11)(25 x-39)>0 \\
\Rightarrow x<\frac{11}{25} .
\end{array}
$$

Also, $x \geqslant-1$, so $-1 \leqslant x<\frac{11}{25}$.