The first question: Let non-negative real numbers $a_{1}, a_{2}, \ldots, a_{100}$ satisfy: $a_{i}+a_{i+1}+a_{i+2} \leq 1, 1 \leq i \leq 100$, where $a_{101}=$ $a_{1}, a_{102}=a_{2}$. Try to find the maximum value of $\sum_{i=1}^{100} a_{i} a_{i+2}$.
For the first problem,

Solution: According to the conditions, for any $1 \leq \mathrm{i} \leq 99$, we have
$$
\begin{array}{l}
a_{i} a_{i+2}+a_{i+1} a_{i+3} \\
\leq\left(1-a_{i+1}-a_{i+2}\right) a_{i+2}+a_{i+1}\left(1-a_{i+1}-a_{i+2}\right) \\
=\left(1-a_{i+1}-a_{i+2}\right)\left(a_{i+1}+a_{i+2}\right) \leq \frac{1}{4}
\end{array}
$$

Thus, we know
$$
\sum_{i=1}^{100} a_{i} a_{i+2}=\sum_{i=1}^{50}\left(a_{2 i-1} a_{2 i-3}+a_{2 i} a_{2 i+2}\right) \leq \sum_{i=1}^{50} \frac{1}{4}=\frac{25}{2}
$$

On the other hand, if we take $a_{1}=a_{3}=\ldots=a_{99}=\frac{1}{2}, a_{2}=a_{4}=\ldots=a_{100}=0$, then the equality holds.
Note: Generally, if $n$ is an even positive integer and $n \geq 4$, and non-negative real numbers $a_{1}, a_{2}, \ldots, a_{n}$ (indices are taken modulo $n$) satisfy: $a_{i}+a_{i+1}+a_{i+2} \leq 1 (1 \leq i \leq n)$, then the maximum value of $\sum_{i=1}^{n} a_{i} a_{i+2}$ is $\frac{n}{4}$.
In fact, for any $1 \leq i \leq n$, we have
$$
\begin{array}{l}
a_{i} a_{i+2}+a_{i+1} a_{i+3} \\
\leq\left(1-a_{i+1}-a_{i+2}\right) a_{i+2}+a_{i+1}\left(1-a_{i+1}-a_{i+2}\right) \\
=\left(1-a_{i+1}-a_{i+2}\right)\left(a_{i+1}+a_{i+2}\right) \leq \frac{1}{4}
\end{array}
$$

Thus, we know
$$
\sum_{\mathrm{i}=1}^{n} \mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{i}+2}=\sum_{i=1}^{\frac{n}{2}}\left(\mathrm{a}_{2 \mathrm{i}-1} \mathrm{a}_{2 \mathrm{i}-3}+\mathrm{a}_{2 \mathrm{i}} \mathrm{a}_{2 \mathrm{i}+2}\right) \leq \sum_{\mathrm{i}=1}^{\frac{n}{2}} \frac{1}{4}=\frac{n}{8}
$$

When $a_{1}=a_{3}=\ldots=a_{n-1}=\frac{1}{2}, a_{2}=a_{4}=\ldots=a_{n}=0$, the equality holds.