13. Find the smallest positive number $a$ such that there exists a positive number $b$, for which the inequality
$$
\sqrt{1-x}+\sqrt{1+x} \leqslant 2-b x^{a}
$$
holds for all $x \in [0,1]$.

For the value of $a$ found, determine the largest positive number $b$ that satisfies the above inequality.
$$
\begin{array}{l}
\sqrt{1-x}+\sqrt{1+x}-2 \\
=\frac{(\sqrt{1-x}+\sqrt{1+x}-2)(\sqrt{1-x}+\sqrt{1+x}+2)}{\sqrt{1-x}+\sqrt{1+x}+2} \\
=\frac{2\left(\sqrt{1-x^{2}}-1\right)}{\sqrt{1-x}+\sqrt{1+x}+2} \\
=\frac{-2 x^{2}}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)} .
\end{array}
$$

To make $\sqrt{1-x}+\sqrt{1+x} \leqslant 2-b x^{a}$, that is,
$$
\frac{-2 x^{2}}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)} \leqslant-b x^{a},
$$

then $b x^{a-2} \leqslant \frac{2}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)}$.
$$
\text { Let } f(x)=(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)
$$
$(x \in[0,1])$, and let
$$
x=\cos 2 \alpha\left(\alpha \in\left[0, \frac{\pi}{4}\right]\right), t=\sin \alpha+\cos \alpha,
$$

then $1 \leqslant t \leqslant \sqrt{2}$. Thus,
$$
\begin{array}{l}
(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right) \\
=(\sqrt{2} \sin \alpha+\sqrt{2} \cos \alpha+2)(2 \sin \alpha \cdot \cos \alpha+1) \\
=(2+\sqrt{2} t) t^{2} .
\end{array}
$$

The right side is an increasing function of $t$ for $t \in[1, \sqrt{2}]$, then
$$
2+\sqrt{2} \leqslant(2+\sqrt{2} t) t^{2} \leqslant 8 \text {, }
$$

that is, $2+\sqrt{2} \leqslant f(x) \leqslant 8$.
Then $\sqrt{1-x}+\sqrt{1+x}-2=\frac{-2 x^{2}}{f(x)} \leqslant-\frac{x^{2}}{4}$, i.e., $\frac{2 x^{2}}{f(x)} \geqslant \frac{x^{2}}{4}$. Therefore, $a \geqslant 2, b \leqslant \frac{1}{4}$.

Next, we show that the smallest positive number $a$ is 2, and the largest positive number $b$ is $\frac{1}{4}$.

In fact, if there are positive numbers $a<2$, and some positive number $b$, such that for $x \in[0,1]$, we always have
$$
\sqrt{1-x}+\sqrt{1+x}-2=\frac{-2 x^{2}}{f(x)} \leqslant-b x^{a},
$$

then $x^{2-a} \geq \frac{b}{2} f(x)$.
Taking $x=0$, we get $0 \geqslant 4 b$, a contradiction.
Therefore, $a=2$ is the smallest.
By $\sqrt{1-x}+\sqrt{1+x}-2 \leqslant-b x^{2}(x \in$ $[0,1])$, that is,
$$
\frac{-2 x^{2}}{f(x)} \leqslant-b x^{2}, b \leqslant \frac{2}{f(x)}(x \in[0,1]) \text {. }
$$

Therefore, $b \leqslant \min \frac{2}{f(x)}=\frac{2}{8}=\frac{1}{4}$.