1. Let real numbers $a_{1}, a_{2}, \cdots, a_{40}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{40}=0 \text {, }
$$

and for $1 \leqslant i \leqslant 40$, we have
$$
\left|a_{i}-a_{i+1}\right| \leqslant 1\left(a_{41}=a_{1}\right) .
$$

Let $a=a_{10}, b=a_{20}, c=a_{30}, d=a_{40}$. 
Find: (1) the maximum value of $a+b+c+d$;
(2) the maximum value of $a b+c d$.
1. (1) It is agreed that the subscripts are understood modulo 40.

According to the conditions, for any integers $k, m$, we have
$a_{m+k} \geqslant a_{m}-|k|$.
Then $0=\sum_{i=1}^{40} a_{i}$
$$
\begin{array}{l}
=\sum_{k=-4}^{5} a_{10+k}+\sum_{k=-4}^{5} a_{00+k}+\sum_{k=-4}^{5} a_{30+k}+\sum_{k=-4}^{5} a_{40+k} \\
\geqslant \sum_{k=-4}^{5}(a-|k|)+\sum_{k=-4}^{5}(b-|k|)+ \\
\sum_{k=-4}^{5}(c-|k|)+\sum_{k=-4}^{5}(d-|k|) \\
\quad=10(a+b+c+d)-4 \times 25 \\
\Rightarrow a+b+c+d \leqslant 10 . \\
\text { Also, let } a_{10 i+k}=\frac{5}{2}-|k|(0 \leqslant i \leqslant 3,-4 \leqslant k \leqslant 5),
\end{array}
$$

At this point, $a_{1}, a_{2}, \cdots, a_{40}$ satisfy the conditions, and
$$
a+b+c+d=4 \times \frac{5}{2}=10 .
$$

Thus, the maximum value of $a+b+c+d$ is 10.
(2) By symmetry, without loss of generality, assume $a b \geqslant c d$.

If $a b \leqslant 0$, then $a b+c d \leqslant 0$.
Now consider the case $a b>0$.
Without loss of generality, assume $a, b>0$ (otherwise, if $a, b \leq 0$, the case is similar).
When $a, b > 0$, if $a+b>10$, by the conclusion of (1), we get $c+d<0$, hence $c, d<0$.
$$
\begin{array}{l}
\text { Also, } c \geqslant b-10, d \geqslant a-10, \text { then } \\
c d \leqslant(10-b)(10-a) .
\end{array}
$$

Note that, $10<a+b \leqslant \frac{25}{2}$.
Thus, $a b+c d$
$$
\begin{array}{l}
\leqslant a b+(10-b)(10-a) \\
=2(a-5)(b-5)+50 \\
\leqslant \frac{1}{2}(a+b-10)^{2}+50 \leqslant \frac{425}{8} .
\end{array}
$$

If $a+b \leqslant 10$, then
$$
a b+c d \leqslant 2 a b \leqslant \frac{(a+b)^{2}}{2} \leqslant 50<\frac{425}{8} \text {. }
$$
(ii) When $c d \leqslant 0$, we have
$$
a b+c d \leqslant a b \leqslant \frac{(a+b)^{2}}{4} \leqslant \frac{625}{16}<\frac{425}{8} \text {. }
$$

On one hand, from cases (i) and (ii), we get
$$
a b+c d \leqslant \frac{425}{8} \text {; }
$$

On the other hand, let
$$
\begin{array}{l}
a_{10+k}=\frac{25}{4}-|k|(-14 \leqslant k \leqslant 5), \\
a_{20+k}=\frac{25}{4}-|k|(-4 \leqslant k \leqslant 15),
\end{array}
$$

At this point, $a_{1}, a_{2}, \cdots, a_{40}$ satisfy the conditions, and
$$
a b+c d=\left(\frac{25}{4}\right)^{2}+\left(-\frac{15}{4}\right)^{2}=\frac{425}{8} .
$$

Thus, the maximum value of $a b+c d$ is $\frac{425}{8}$.