$(1)$ Set $s_1 = \frac12 a_5 + a_6 + a_7 + \cdots + a_{14} + \frac12 a_{15}.$ Let $s_2 = \frac12 a_{15} + a_{16} + a_{17} + \cdots + a_{24} + \frac12 a_{25}.$ Define $s_3, s_4$ in a similar manner. Observe that $s_1 \ge 10a_{10} - 2 \cdot 1 - 2\cdot 2 - 2 \cdot 3 - 2 \cdot 4 - 5 = 10 a_{10} - 25.$ Summing this with three similar inequalities for $s_2, s_3, s_4$, we obtain: $$0 = s_1 + s_2 + s_3 + s_4 \ge 10(a_{10} + a_{20} + a_{30} + a_{40}) - 100,$$ which yields $a_{10} + a_{20} + a_{30} + a_{40} \le 10$. This is attained when $a_{10} = a_{20} = a_{30} = a_{40} = 2.5$ and $a_5 = a_{15} = a_{25} = a_{35} = -2.5,$ so the answer is $\boxed{10}.$ $(2)$ Let $x = a_{10} + a_{20}$ and $y = a_{30} + a_{40}.$ Then it's clear that $a_{10} \cdot a_{20} + a_{30} \cdot a_{40} \le \frac{x^2 + y^2}{4}.$ If $x, y$ are both nonnegative, then we have from $(1)$ that $\frac{x^2 + y^2}{4} \le \frac{(x+y)^2}{4} \le 25.$ If $x, y$ are both nonpositive, then negate all the $a_i$'s and continue as in the previous case. So WLOG assume that $x > 0 > y.$ Notice that $a_{10} - a_{40} \le 10$ and $a_{20} - a_{30} \le 10$ from the condition, and so $x-y \le 20.$ Claim. $x \le 12.5.$ Proof. Suppose that $a_{10} + a_{20} > 12.5,$ for contradiction. Let $t = a_{10}, u = a_{20}.$ Then, notice that $\frac12 a_{15} + a_{14} + a_{13} + a_{12} + \cdots + a_1 + a_{40} + a_{39} + \cdots + a_{36} + \frac12 a_{35} \ge 20t - 125.$ Analogously, $\frac12 a_{15} + a_{16} + a_{17} + a_{18} + \cdots + a_{34} + \frac12 a_{35} \ge 20u - 125.$ Summing, we get that $0 \ge 20 (t+u) - 250$, which implies the claim. $\blacksquare$ Analogously, we get that $y \ge -12.5$. From $x>0>y, x \le 12.5, y \ge -12.5,$ and $x - y \le 20$, it becomes apparent that $a_{10} \cdot a_{20} + a_{30} \cdot a_{40} \le \frac{x^2 + y^2}{4} \le 6.25^2 + 3.75^2.$ Indeed, this is attainable when $a_{10} = a_{20} = 6.25$ and $a_{30} = a_{40} = -3.75,$ so our answer is $\boxed{6.25^2 + 3.75^2}.$ $\square$

$$6.25^2 + 3.75^2=\frac{425}{8} .$$Let $a_1,a_2,a_3,\cdots,a_{40}$ be reals such that $a_1+a_2+a_3+\cdots+a_{40}=0 $ and $|a_i-a_{i+1}|\leq 1$ $(i=1,2,3,\cdots,40), a_{41}=a_1.$ Then $$a_{10}+a_{20}+a_{30}+a_{40}\leq 10,$$$$a_{10}a_{20}+a_{30}a_{40}\leq\frac{425}{8} .$$

The answers are $10$ and $\tfrac{425}{8}=53.125$ respectively. We prove part 1 and 2 separately. Extend the indices modulo $40$. To prove the bound in part 1, write the following ten inequalities \begin{align*} a_5 &\geq a_{10}-5 \\ a_6 &\geq a_{10}-4 \\ a_7 &\geq a_{10}-3 \\\ a_8 &\geq a_{10}-2 \\ a_9 &\geq a_{10}-1 \\ a_{10} &\geq a_{10} \\ a_{11} &\geq a_{10}-1 \\ a_{12} &\geq a_{10}-2 \\ a_{13} &\geq a_{10}-3 \\ a_{14} &\geq a_{10}-4 \end{align*}Summing up gives $a_5+a_6+\hdots + a_{14}\geq 10a_{10}-25$. Now we write an analogous inequalities for $a_{20}, a_{30}, a_{40}$. \begin{align*} a_5+a_6+\hdots + a_{14}&\geq 10a_{10}-25 \\ a_{15}+a_{16}+\hdots + a_{24}&\geq 10a_{20}-25 \\ a_{25}+a_{26}+\hdots + a_{34}&\geq 10a_{30}-25 \\ a_{35}+a_{36}+\hdots + a_{44}&\geq 10a_{40}-25 \\ \end{align*}Adding all inequalities gives $0\geq 10(a_{10}+a_{20}+a_{30}+a_{40})-100$ which is the desired bound. It's possible that all inequalities above are equalities, say $a_{10}=a_{20}=a_{30}=a_{40}=2.5$ while $a_5=a_{15}=a_{25}=a_{35}=-2.5$. For part 2, we prove the following claim. Claim: $a_{10}+a_{20}\leq 12.5$. Proof: Very similar to above. Shift the indices to $a_{15}+a_{25}$. Write the following twenty inequalities. \begin{align*} a_1&\geq a_{15}-14 \\ a_2&\geq a_{15}-13 \\ &\ \vdots \\ a_{14}&\geq a_{15}-1\\ a_{15}&\geq a_{15} \\ a_{16}&\geq a_{15}-1 \\ a_{17}&\geq a_{15}-2 \\ a_{18}&\geq a_{15}-3 \\ a_{19}&\geq a_{15}-4 \\ a_{20}&\geq a_{15}-5 \\ \end{align*}Summing up gives $a_1+a_2+\hdots+a_{20}\geq 20a_{15}-120$. For bound around $a_{25}$, we have the following twenty inequalities. \begin{align*} a_{21}&\geq a_{25}-4 \\ a_{22}&\geq a_{25}-3 \\ a_{23}&\geq a_{25}-2 \\ a_{24}&\geq a_{25}-1 \\ a_{25}&\geq a_{25} \\ a_{26}&\geq a_{25}-1 \\ a_{27}&\geq a_{25}-2 \\ &\ \vdots \\ a_{40}&\geq a_{25}-15 \end{align*}Summing up gives $a_{21}+a_{22}+\hdots+a_{40}\geq 20a_{25}-130$. Thus $20(a_{15}+a_{25})-250\geq 0$, which gives the desired claim. $\blacksquare$ Back to the main problem, let $x=a_{10}+a_{20}$ and $y=a_{30}+a_{40}$. We have the following estimates. $$M = a_{10}a_{20}+a_{30}a_{40}\leq \frac{x^2+y^2}{4}.$$If $x,y\geq 0$, then since $x+y\leq 10$, we have the RHS must not exceed $25$ which is not sharp. Similarly $x,y\leq 0$ does not give the maximum. Therefore assume that $x>0, y<0$. Observe that $$|x-y| \leq |a_{10}-a_{0}| + |a_{30}-a_{20}| \leq 20$$thus $x\leq 12.5$ and $y\geq -7.5$. Hence $M\leq 6.25^2+3.75^2 = \tfrac{425}{8}$ as desired. The equality holds when $a_{10}=a_{20}=6.25$, $a_{30}=a_{40}=-3.75$, $a_{15}=11.25$ and $a_{35}=-8.75$.

im stumped

same here