$$
\begin{array}{c}
\text { II. (50 points) }(1) \text { For } 0 \leqslant x \leqslant 1, \text { find the range of the function } \\
h(x)=(\sqrt{1+x}+\sqrt{1-x}+2) . \\
\left(\sqrt{1-x^{2}}+1\right)
\end{array}
$$

(2) Prove: For $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequality
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{a}}{\beta}
$$

holds for the smallest positive number $\alpha=2$. And find the smallest positive number $\beta$ at this time.
$$
\begin{array}{l}
=(1) 00$, the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{a}}{\beta}(x \in[0,1])$ does not hold. Conversely, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{0}}{\beta}$, which means $x^{2 \cdots} \geqslant \frac{h(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get
$$
0 \geqslant \frac{h(0)}{2 \beta} \text {. }
$$

Estimate $h(0)=8$.
This is impossible. This indicates that $\alpha=2$ is the smallest positive number that satisfies the condition.

Find the smallest $\beta>0$ such that the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{2}}{\beta}(x \in[0,1])$, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{2}}{\beta}$
holds, which is equivalent to finding
$$
\beta=\max _{0 \leqslant x \leqslant 1} \frac{1}{2} h(x) \text {. }
$$
$\because$ For any real numbers $u, v$, we have
$$
\sqrt{u}+\sqrt{v} \leqslant \sqrt{2} \bar{u}+v) \text {. }
$$

Let $u=1+x, v=1-x$, we get
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2 .
$$

Thus, when $x \in[0,1]$,
$$
\begin{aligned}
\frac{1}{2} h(x) & =\frac{1}{2}(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right) \\
& \leqslant 2\left(\sqrt{1-x^{2}}+1\right) \leqslant 4 .
\end{aligned}
$$

And $\frac{1}{2} h(0)=4$. Therefore, the maximum value of the function $\frac{1}{2} h(x)$ on $[0,1]$ is 4, i.e., the smallest positive number $\beta$ that satisfies condition (2) is 4.
$$