1. The solution set of the inequality $x^{3}+\left(1-x^{2}\right)^{\frac{3}{2}} \geqslant 1$ is
1. $\{0,1\}$.
Let $\sqrt{1-x^{2}}=y$, then the inequality becomes
$$
x^{2}+y^{2}=1, x^{3}+y^{3} \geqslant 1 \text {. }
$$

Thus, $0 \leqslant x^{3}+y^{3}-x^{2}-y^{2}$
$$
\begin{array}{l}
=x^{2}(x-1)+y^{2}(y-1) \\
=\left(1-y^{2}\right)(x-1)+\left(1-x^{2}\right)(y-1) \\
=-\left(y^{2}-1\right)(x-1)-\left(x^{2}-1\right)(y-1) \\
=-(1-x)(1-y)(x+y+2) .
\end{array}
$$

Since $1=x^{2}+y^{2} \geqslant x^{2}$, we have $|x| \leqslant 1$.
Similarly, $|y| \leqslant 1$.
Therefore, $1 \pm x \geqslant 0,1 \pm y \geqslant 0, x+y+2 \geqslant 0$.
If $x+y+2=0$, then $x+1=y+1=0$, which does not satisfy $x^{2}+y^{2}=1$. Therefore, $x+y+2>0$.
Thus, the inequality becomes $(1-x)(1-y) \leqslant 0$.
But $1-x \geqslant 0,1-y \geqslant 0$, so
$$
(1-x)(1-y)=0 \text {. }
$$

Solving gives $(x, y)=(1,0),(0,1)$.
Upon verification, $x=0$ or 1 are both solutions to the original inequality.
Therefore, the solution set of the original inequality is $\{0,1\}$.