According to the problem, we can set $l_{A B}: y=k x+1(k>0)$, then $l_{B C}: y=-\frac{1}{k} x+1$. From $\left\{\begin{array}{l}x^{2}+a^{2} y^{2}=a^{2}, \\ y=k x+1,\end{array}\right.$ we get $\left(1+k^{2} a^{2}\right) x^{2}+2 a^{2} k x=0$, solving for $x_{A}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$, then $|A B|=$ $\sqrt{1+k^{2}}\left|\frac{-2 a^{2} k}{1+a^{2} k^{2}}\right|=\sqrt{1+k^{2}} \frac{2 a^{2} k}{1+a^{2} k^{2}}$. Similarly, $|B C|=\sqrt{1+\frac{1}{k^{2}}}\left|\frac{-2 a^{2} \cdot\left(-\frac{1}{k}\right)}{1+a^{2} \cdot \frac{1}{k^{2}}}\right|=$ $\frac{\sqrt{1+k^{2}}}{k} \cdot \frac{2 a^{2} k}{a^{2}+k^{2}}$.

Since $\triangle A B C$ is an isosceles right triangle, we have $|A B|=$ $|B C|$, so
$$
\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2} k^{2}}=\frac{\sqrt{1+k^{2}}}{k} \cdot \frac{2 a^{2} k}{a^{2}+k^{2}},
$$

Simplifying, we get $k^{3}-a^{2} k^{2}+a^{2} k-1=0$, then $(k-1)\left[k^{2}+\left(1-a^{2}\right) k+1\right]=0$.
Since there is only one isosceles right triangle that meets the conditions, the above equation has only one real root. $k=1$, so the equation $k^{2}+\left(1-a^{2}\right) k+1=0$ has no real roots or has two equal real roots $k=1$. Thus, $\Delta=\left(1-a^{2}\right)^{2}-4=a^{4}-2 a^{2}-31$, solving for $1<a<\sqrt{3}$, when $\Delta=0$, solving for $a=\sqrt{3}$, at this time the equation is $k^{2}-2 k+1=0$, which has two equal real roots $k=1$. In summary, the range of $a$ is $1<a \leqslant \sqrt{3}$.
For the ellipse $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$, construct an isosceles right triangle $\triangle A B C$ inscribed in the ellipse with the right angle vertex at the endpoint $B(0,1)$ of the minor axis. If there is exactly one such triangle, find the range of the real number $a$.