Third question: For all real numbers $x_{1}, x_{2}, \ldots, x_{60} \in[-1,1]$, find the maximum value of $\sum_{i=1}^{60} x_{i}^{2}\left(x_{i+1}-x_{i-1}\right)$, where $x_{0}=x_{60}, x_{61}=x_{1}$.
Second question, Solution method one: First, take $a_{1}=a_{n}=\frac{1}{2}, a_{2}=a_{3}=\ldots=a_{n-1}=0$, then $\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}=\frac{n-1}{4}$.
Next, we prove by induction on $n$: $\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n}}(\mathrm{j}-\mathrm{i}) \mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{j}} \leq \frac{\mathrm{n}-1}{4}$.
When $n=2$, $(2-1) a_{1} a_{2} \leq \frac{(a_{1}+a_{2})^{2}}{4}=\frac{1}{4}$, the conclusion holds.
Assume the conclusion holds for $n$, now consider the case for $n+1$.
By the induction hypothesis, we know:
$$
\sum_{1 \leq i<1 \leq n-1}(j-i) a_{i} a_{j}+\sum_{i=1}^{n-1}(n-i) a_{i}(a_{n}+a_{n+1}) \leq \frac{n-1}{4}
$$

And according to the condition, we have:
$$
\begin{array}{l}
\sum_{1 \leq i<i \leq n+1}(j-i) a_{i} a_{j}-\left(\sum_{1 \leq i<i \leq n-1}(j-i) a_{i} a_{j}+\sum_{i=1}^{n-1}(n-i) a_{i}(a_{n}+a_{n+1})\right) \\
=[(n+1)-n] a_{n} a_{n+1}+\sum_{i=1}^{n-1}[(n+1-i)-(n-i)] a_{i} a_{n+1} \\
=a_{n+1}\left(\sum_{i=1}^{n} a_{i}\right) \leq \frac{\left(\sum_{i=1}^{n+1} a_{i}\right)^{2}}{4}=\frac{1}{4}
\end{array}
$$

Thus, $\sum_{1 \leq i<j \leq n+1}(j-i) a_{i} a_{j} \leq \frac{n-1}{4}+\frac{1}{4}=\frac{n}{4}$.
In conclusion, the maximum value of $\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}$ is $\frac{n-1}{4}$.
Solution method two: First, take $a_{1}=a_{n}=\frac{1}{2}, a_{2}=a_{3}=\ldots=a_{n-1}=0$, then $\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}=\frac{n-1}{4}$.
On the other hand, let $S_{k}=\sum_{i=1}^{k} a_{i}$, then $S_{n}=1$, and
$$
\begin{array}{l}
\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}=\frac{\sum_{i=1}^{n} a_{i}^{2}+2 \sum_{1 \leq i<1 \leq n} j a_{i} a_{j}}{2}-\frac{\sum_{i=1}^{n} i a_{i}^{2}+2 \sum_{1 \leq i<j \leq n} a_{i} a_{i}}{2} \\
=\frac{n\left(\sum_{i=1}^{n} a_{i}\right)^{2}-\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} a_{i}\right)^{2}}{2}-\frac{\sum_{k=1}^{n}\left(\sum_{i=k}^{n} a_{i}\right)^{2}}{2} \\
=\frac{n S_{n}^{2}-\sum_{k=1}^{n-1} s_{k}^{2}}{2}-\frac{S_{n}^{2}+\sum_{k=1}^{n-1}\left(S_{n}-S_{k}\right)^{2}}{2} \\
=\frac{(n-1) S_{n}^{2}-\sum_{k=1}^{n-1}\left(s_{k}^{2}+\left(S_{n}-S_{k}\right)^{2}\right)}{2} \\
\leq \frac{(n-1) S_{n}^{2}-\sum_{k=1}^{n-1} \frac{\left[s_{k}+\left(s_{n}-s_{k}\right)\right]^{2}}{2}}{2}=\frac{n-1}{4}
\end{array}
$$

Third question,
Solution: Transform to get:
$$
\begin{array}{l}
\sum_{i=1}^{60} x_{i}^{2}\left(x_{i+1}-x_{i-1}\right)=\sum_{i=1}^{60} x_{i}^{2} x_{i+1}-\sum_{i=1}^{60} x_{i}^{2} x_{i-1}=\sum_{j=1}^{60} x_{i} x_{i+1}\left(x_{i}-x_{i+1}\right) \\
=\frac{\sum_{i=1}^{60}\left(\left(x_{i}^{3}-x_{i+1}^{3}\right)-\left(x_{i}-x_{i+1}\right)^{3}\right)}{3}=\frac{\sum_{i=1}^{60}\left(x_{i+1}-x_{i}\right)^{3}}{3}
\end{array}
$$

Notice that, when $a \in[-2,2]$, $a^{3}-(3 a+2)=(a+1)^{2}(a-2) \leq 0$, hence $a^{3} \leq 3 a+2$. For any $1 \leq i \leq 60$, we have $x_{i+1}-x_{i} \in[-2,2]$. Therefore, we know:
$$
\sum_{i=1}^{60} x_{i}^{2}\left(x_{i+1}-x_{i-1}\right)=\frac{\sum_{i=1}^{60}\left(x_{i+1}-x_{i}\right)^{3}}{3} \leq \frac{\sum_{i=1}^{60}\left[3\left(x_{i+1}-x_{i}\right)+2\right]}{3}=40
$$

When $\left\{\begin{array}{c}x_{3 k+1}=1 \\ x_{3 k+2}=0 \\ x_{3 k+3}=-1\end{array}(0 \leq k \leq 19)\right.$, the equality holds.