1. Find the largest real number $\theta(\theta<\pi)$ such that
$$
\prod_{k=0}^{10} \cos 2^{k} \theta \neq 0 \text {, and } \prod_{k=0}^{10}\left(1+\frac{1}{\cos 2^{k} \theta}\right)=1
$$
Let $z=\mathrm{e}^{\frac{10}{2}} \neq \pm 1$. Then
$$
\begin{array}{l}
\prod_{k=0}^{10}\left(1+\frac{1}{\cos 2^{k} \theta}\right) \\
=\frac{z+z^{-1}}{z^{2^{11}}+z^{-2^{11}}} \cdot \frac{z^{2^{11}}-z^{-2^{11}}}{z-z^{-1}} \\
=\frac{\tan 2^{10} \theta}{\tan \frac{\theta}{2}}=1 .
\end{array}
$$

Given that the period of $\tan x$ is $\pi$, and it is injective within any given period, it is easy to see that
$$
\frac{\theta}{2}+l \pi=1024 \theta(l \in \mathbf{Z}) .
$$

Therefore, $\theta=\frac{2 l \pi}{2047}$.
When $l=1023$, $\theta$ reaches its maximum value $\frac{2046 \pi}{2047}$.