1. Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy $9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015)$. Find the maximum value of $\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$.
Let $P=\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$,
Since $a_{i}>\frac{11}{9} a_{i+1}^{2} \Rightarrow a_{i}-a_{i+1}^{2}>\frac{11}{9} a_{i+1}^{2}-a_{i+1}^{2}=\frac{2}{9} a_{i+1}^{2} \geqslant 0$,
If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$; If $a_{2016}-a_{1}^{2}>0$, let $a_{2017}=a_{1}$,
then $P^{\frac{1}{2016}} \leqslant \frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right)=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i+1}^{2}\right)$
$=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016}\left[-\left(a_{i}-\frac{1}{2}\right)^{2}+\frac{1}{4}\right]$
$\leqslant \frac{1}{2016} \sum_{i=1}^{2016} \frac{1}{4}=\frac{1}{4}$, equality holds when $a_{i}=\frac{1}{2}, i=1,2, \cdots, 2016$.
At this point, $9 a_{i}=\frac{9}{2}>\frac{11}{4}=11 a_{i+1}^{2}$ is clearly satisfied, so the maximum value of $P^{\frac{1}{2016}}$ is $\frac{1}{4}$.
In summary, the maximum value of $\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$ is $\frac{1}{4^{2016}}$.