Question 92, Solution: According to the conditions, $|\vec{a}|=|\vec{b}|, \vec{a} \perp \vec{b}$, let $\vec{a}=(x, 0), \vec{b}=(0, x)$, then $\vec{c}=(\lambda x, \mu x)$. Therefore, we have:
$$
\left\{\begin{array}{c}
|\vec{c}|=(\lambda x)^{2}+(\mu x)^{2}=1 \\
(\vec{a}+\vec{b}) \cdot \vec{c}=\lambda x^{2}+\mu x^{2}=1
\end{array}\right.
$$

Let $\lambda x=\cos \alpha, \mu x=\sin \alpha$, then $\left\{\begin{array}{c}x=\frac{1}{\sin \alpha+\cos \alpha} \\ \lambda=\cos \alpha(\sin \alpha+\cos \alpha) \\ \mu=\sin \alpha(\sin \alpha+\cos \alpha)\end{array}, \vec{c}=(\cos \alpha, \sin \alpha), \vec{a}=\right.$ $\left(\frac{1}{\sin \alpha+\cos \alpha}, 0\right)$. Therefore:
$$
\begin{array}{l}
\frac{|\alpha-\tau|}{|1+\mu-\lambda|}=\sqrt{\frac{\left(\frac{1}{\sin \alpha+\cos \alpha} \cos \alpha\right)^{2}+(0-\sin \alpha)^{2}}{(1+(\sin \alpha-\cos \alpha)(\sin \alpha+\cos \alpha))^{2}}}=\sqrt{\frac{1}{2(\sin \alpha(\sin \alpha+\cos \alpha))^{2}}} \\
=\sqrt{\frac{2}{(1+\sin 2 \alpha-\cos 2 \alpha)^{2}}} \geq \sqrt{\frac{2}{(1+\sqrt{2})^{2}}}=2-\sqrt{2}
\end{array}
$$

In summary, the minimum possible value of $\frac{|\alpha-\tau|}{|1+\mu-\lambda|}$ is $2-\sqrt{2}$.
Question 92, Given planar vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy: $|\vec{a}|=|\vec{b}|, \vec{c}=\lambda \vec{a}+\mu \vec{b},|\vec{c}|=1+\vec{a} \cdot \vec{b}=$ $(\vec{a}+\vec{b}) \cdot \vec{c}=1$, try to find the minimum possible value of $\frac{|\vec{a}-\vec{c}|}{|1+\mu|}$.