The first question: Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy
$$
\begin{array}{l}
9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015) \text {. Find } \\
\quad\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)
\end{array}
$$

the maximum value.
Solution 1 First prove a lemma
Lemma $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in \mathbf{R}_{+}$, $n \geqslant 2, a_{i}>b_{i}(1 \leqslant i \leqslant n)$. Then
$\prod_{i=1}^{n}\left(a_{i}-b_{i}\right) \leqslant\left(\sqrt[n]{\prod_{i=1}^{n} a_{i}}-\sqrt[n]{\prod_{i=1}^{n} b_{i}}\right)^{n}$.
Proof (1) Prove the case when $n=2^{k}\left(k \in \mathbf{Z}_{+}\right)$. Use mathematical induction on $k$.
When $k=1$,
$\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right) \leqslant\left(\sqrt{a_{1} a_{2}}-\sqrt{b_{1} b_{2}}\right)^{2}$
$\Leftrightarrow a_{1} b_{2}+a_{2} b_{1}-2 \sqrt{a_{1} b_{2} a_{2} b_{1}} \geqslant 0$
$\Leftrightarrow\left(\sqrt{a_{1} b_{2}}-\sqrt{a_{2} b_{1}}\right)^{2} \geqslant 0$.
The proposition holds.
Assume the proposition holds when $k=s$,
Then when $k=s+1$,
$\prod_{i=1}^{2^{s+1}}\left(a_{i}-b_{i}\right)$
$=\left[\prod_{i=1}^{2^{s}}\left(a_{i}-b_{i}\right)\right]\left[\prod_{i=1}^{2^{s}}\left(a_{i+2^{s}}-b_{i+2^{s}}\right)\right]$
$\leqslant\left[\left(\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} a_{i}}-\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} b_{i}}\right)\right.$
$\left.\left(\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} a_{i+2^{s}}}-\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} b_{i+2^{s}}}\right)\right]^{2^{s}}$
$\leqslant\left(\sqrt{\prod_{i=1}^{2^{s+1}} a_{i}}-\sqrt{\prod_{i=1}^{2^{s+1}} b_{i}}\right)^{2^{s+1}}$.
Thus, the proposition holds in this case.
(2) Prove: If the proposition holds when $n=r+1$,
then the proposition also holds when $n=r$
$\left(r \geqslant 2, r \in \mathbf{Z}_{+}\right)$.
Let $A=\sqrt[r]{\prod_{i=1}^{r} a_{i}}, B=\sqrt[r]{\prod_{i=1}^{r} b_{i}}$,
$a_{i}^{*}=a_{i}, b_{i}^{*}=b_{i}(1 \leqslant i \leqslant r)$,
$a_{r+1}=A, b_{r+1}=B$.
Then $A>B$.
By the induction hypothesis,
$\prod_{i=1}^{r+1}\left(a_{i}^{*}-b_{i}^{*}\right)$
$\leqslant\left(\sqrt{\prod_{i=1}^{r+1} a_{i}^{*}}-\sqrt{\prod_{i=1}^{r+1} b_{i}^{*}}\right)^{r+1}$
$=(A-B)^{r+1}$
$\Rightarrow \prod_{i=1}^{r}\left(a_{i}-b_{i}\right)=\frac{\prod_{i=1}^{r+1}\left(a_{i}^{*}-b_{i}^{*}\right)}{A-B} \leqslant(A-B)^{r}$.
In summary, the lemma is proved.
For any $i(1 \leqslant i \leqslant 2015)$, by the given conditions,
$9 a_{i}>11 a_{i+1}^{2} \geqslant 9 a_{i+1}^{2} \geqslant 0 \Rightarrow a_{i}-a_{i+1}^{2}>0$
If $a_{2016}-a_{1}^{2} \leqslant 0$, then
$M=\prod_{i=2010}^{2016}\left(a_{i}-a_{i+1}^{2}\right) \leqslant 0\left(a_{2017}=a_{1}\right)$.
Therefore, we only need to consider the case $a_{2016}-a_{1}^{2}>0$,
in which case,
$\prod_{i=1}^{2016} a_{i}>\prod_{i=1}^{2016} a_{i+1}^{2} \Rightarrow S=\sqrt{\prod_{i=1}^{2016} a_{i}}$.
If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$.
Below, consider the case $a_{2016}-a_{1}^{2}>0$.
Let $m_{i}=\frac{a_{i}}{a_{i+1}^{2}}(i=1,2, \cdots, 2015)$,
$m_{2016}=\frac{a_{2016}}{a_{1}^{2}}$,

Then, $\prod_{i=1}^{2016} m_{i}=\prod_{i=1}^{2016} \frac{1}{a_{i}}$.
Also, $m_{i}>\frac{11}{9}(i=1,2, \cdots, 2015)$, so
$P=\prod_{i=1}^{2016}\left[\left(m_{i}-1\right) a_{i}^{2}\right]$
$=\left[\prod_{i=1}^{2016}\left(m_{i}-1\right)\right]\left(\prod_{i=1}^{2016} a_{i}^{2}\right)$
$=\left[\prod_{i=1}^{2016}\left(m_{i}-1\right)\right]\left(\prod_{i=1}^{2016} \frac{1}{m_{i}^{2}}\right)$
$=\prod_{i=1}^{2016} \frac{m_{i}-1}{m_{i}^{2}}$.
Notice that,
$\frac{m_{i}-1}{m_{i}^{2}}=-\left(\frac{1}{m_{i}}-\frac{1}{2}\right)^{2}+\frac{1}{4} \leqslant \frac{1}{4}$.
Thus, $P=\prod_{i=1}^{2016} \frac{m_{i}-1}{m_{i}^{2}} \leqslant \frac{1}{4^{2016}}$.
The equality holds if and only if
$m_{i}=2(i=1,2, \cdots, 2016)$,

which means $a_{i}=2 a_{i+1}^{2}(i=1,2, \cdots, 2015), a_{2016}=2 a_{1}^{2}$.
Solving this, we get $a_{i}=\frac{1}{2}(i=1,2, \cdots, 2016)$.
Upon verification, this satisfies the given inequalities, indicating that the equality can hold.
In summary, the maximum value sought is $\frac{1}{4^{2016}}$.