For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
(i) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
(ii) There exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, such that $\left[(1-x) \sin \theta-\sqrt{a-x^{2} \cos ^{2} \theta}\right]^{2}+$ $\left[x \cos \theta-\sqrt{a-(1-x)^{2} \sin ^{2} \theta}\right]^{2} \leqslant a$.
From (i) we get $\sqrt{a}>\frac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}$.
(1)
(ii) is equivalent to: there exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, satisfying
$$
2 \sin \theta \cos \theta\left[(1-x) \sqrt{\frac{a}{\cos ^{2} \theta}-x^{2}}+x \sqrt{\frac{a}{\sin ^{2} \theta}-(1-x)^{2}}\right] \geqslant a .
$$

First, we prove a lemma: Let $01, p^{2}+q^{2} \leqslant 1, f(x)=$ $(1-x) \sqrt{p^{2}-x^{2}}+x \sqrt{q^{2}-(1-x)^{2}}(1-q \leqslant x \leqslant p)$. Then when $\sqrt{p^{2}-x^{2}}=$ $\sqrt{q^{2}-(1-x)^{2}}$, i.e., $x=\frac{p^{2}-q^{2}+1}{2} \in[1-q, p]$, $f(x)$ reaches its maximum value.

Since $1-q \leqslant x \leqslant p$, we can let $x=p \sin \alpha, 1-x=q \sin \beta, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, $0<\alpha+\beta<\pi$. Thus, $f(x)=p q(\sin \beta \cos \alpha+\sin \alpha \cos \beta)-p q \sin (\alpha+\beta)$. And $\cos (\alpha+$ $\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta)=\frac{\sqrt{p^{2}-x^{2}} \cdot \sqrt{q^{2}-(1-x)^{2}}-x(1-x)}{p q}=$ $\frac{p^{2}+q^{2}-1-\left(\sqrt{p^{2}-x^{2}}-\sqrt{q^{2}-(1-x)^{2}}\right)^{2}}{2 p q} \leqslant 0$, thus $\frac{\pi}{2} \leqslant \alpha+\beta<\pi$. At the same time, when and only when $\sqrt{p^{2}-x^{2}}=\sqrt{q^{2}-(1-x)^{2}}$, i.e., $x=\frac{1}{2}\left(p^{2}-q^{2}+1\right) \in[1-q, p]$, $\cos (\alpha+\beta)$ reaches its maximum value $\frac{p^{2}+q^{2}-1}{2 p q} \leqslant 0$. Because the sine function is monotonically decreasing in $\left[\frac{\pi}{2}, \pi\right]$, $f(x)=p q \sin (\alpha+\beta)$ also reaches its maximum value when and only when $x=\frac{1}{2}\left(p^{2}-q^{2}+1\right)$. The lemma is proved.

By the lemma, when $\frac{a}{\sin ^{2} \theta}+\frac{a}{\cos ^{2} \theta} \leqslant 1$, when and only when $\sqrt{\frac{a}{\cos ^{2} \theta}-x^{2}}=$ $\sqrt{\frac{a}{\sin ^{2} \theta}-(1-x)^{2}}$, i.e., $x=\frac{1}{2}\left(\frac{a}{\cos ^{2} \theta}-\frac{a}{\sin ^{2} \theta}+1\right) \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, it reaches the maximum value $2 \sin \theta \cos \theta \sqrt{\frac{a}{\cos ^{2} \theta}-\frac{1}{4}\left(\frac{a}{\cos ^{2} \theta}-\frac{a}{\sin ^{2} \theta}+1\right)^{2}}$. From (2), the smallest $a$ that satisfies the condition and (1) is:
$$
2 \sin \theta \cos \theta \sqrt{\frac{a}{\cos ^{2} \theta}-\frac{1}{4}\left(\frac{a}{\cos ^{2} \theta}-\frac{a}{\sin ^{2} \theta}+1\right)^{2}} \geqslant a,
$$

i.e., $\left(1-3 \sin ^{2} \theta \cos ^{2} \theta\right) a^{2}-2 \sin ^{2} \theta \cos ^{2} \theta a+\sin ^{4} \theta \cos ^{4} \theta \leqslant 0$. Solving this, we get
$$
\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} \leqslant a \leqslant \frac{\sin ^{2} \theta \cos ^{2} \theta}{1-\sqrt{3} \sin \theta \cos \theta}
$$

Since $\frac{\sin ^{2} \theta \cos ^{2} \theta}{(\sin \theta+\cos \theta)^{2}}<\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta}$, we have $\frac{a}{\cos ^{2} \theta}+\frac{a}{\sin ^{2} \theta}=\frac{a}{\cos ^{2} \theta \cos ^{2} \theta}=$ $\frac{1}{1+\sqrt{3} \sin \theta \cos \theta}<1$. Therefore, when $a=\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta}$, it satisfies (1), hence this is the solution.

Note: There are two points worth noting in the above analysis: 1. The problem structure is complex, so we first prove a more general case, the lemma; 2. Notice that $\sin (\alpha+\beta)$ and $\cos (\alpha+\beta)$ have the same monotonicity in $\alpha+\beta \in\left[\frac{\pi}{2}, \pi\right]$, thus we find the maximum value of $\cos (\alpha+\beta)$ to find the maximum value of $\sin (\alpha+\beta)$.