1. ĐK: $x\ge-3$.
Đặt: $u=\sqrt{x+3};v=\sqrt[3]{x};u\ge0$.
Hệ trở thành: $\left\{\begin{array}{l}u-v=1\\u^2-v^3=3\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}u=v+1\\(v+1)^2-v^3=3\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}u=2\\v=1\end{array}\right.\\\left\{\begin{array}{l}u=1+\sqrt2\\v=\sqrt2\end{array}\right.\\\left\{\begin{array}{l}u=1-\sqrt2\\v=-\sqrt2\end{array}\right.\end{array}\right.$
Từ đó suy ra: $x\in\{1;2\sqrt2\}$.