Let $a_1, a_2, \ldots, a_{41} \in \mathbb{R}$ such that $a_{41} = a_1$, $\sum_{i=1}^{40} a_i = 0$, and for any $i = 1, 2, \ldots, 40$, $|a_i - a_{i+1}| \leq 1$. We aim to determine the greatest possible values of:
1. $a_{10} + a_{20} + a_{30} + a_{40}$
2. $a_{10} \cdot a_{20} + a_{30} \cdot a_{40}$

### Part 1
Let $s_1 = \frac{1}{2} a_5 + a_6 + a_7 + \cdots + a_{14} + \frac{1}{2} a_{15}$. Define $s_2, s_3, s_4$ similarly. Observe that:
$$s_1 \geq 10a_{10} - 2 \cdot 1 - 2 \cdot 2 - 2 \cdot 3 - 2 \cdot 4 - 5 = 10a_{10} - 25.$$

Summing this with three similar inequalities for $s_2, s_3, s_4$, we obtain:
$$0 = s_1 + s_2 + s_3 + s_4 \geq 10(a_{10} + a_{20} + a_{30} + a_{40}) - 100,$$
which yields:
$$a_{10} + a_{20} + a_{30} + a_{40} \leq 10.$$

This is attained when $a_{10} = a_{20} = a_{30} = a_{40} = 2.5$ and $a_5 = a_{15} = a_{25} = a_{35} = -2.5$. Therefore, the greatest possible value of $a_{10} + a_{20} + a_{30} + a_{40}$ is:
$$\boxed{10}.$$

### Part 2
Let $x = a_{10} + a_{20}$ and $y = a_{30} + a_{40}$. Then:
$$a_{10} \cdot a_{20} + a_{30} \cdot a_{40} \leq \frac{x^2 + y^2}{4}.$$

From Part 1, we know $x + y \leq 10$. If $x$ and $y$ are both nonnegative, then:
$$\frac{x^2 + y^2}{4} \leq \frac{(x+y)^2}{4} \leq 25.$$

If $x$ and $y$ are both nonpositive, negate all $a_i$'s and continue as in the previous case.

Assume $x > 0 > y$. Notice that $a_{10} - a_{40} \leq 10$ and $a_{20} - a_{30} \leq 10$, so $x - y \leq 20$.

Claim: $x \leq 12.5$.

Proof: Suppose $a_{10} + a_{20} > 12.5$. Let $t = a_{10}$ and $u = a_{20}$. Then:
$$\frac{1}{2} a_{15} + a_{14} + a_{13} + \cdots + a_1 + a_{40} + a_{39} + \cdots + a_{36} + \frac{1}{2} a_{35} \geq 20t - 125,$$
and similarly:
$$\frac{1}{2} a_{15} + a_{16} + a_{17} + \cdots + a_{34} + \frac{1}{2} a_{35} \geq 20u - 125.$$

Summing these, we get:
$$0 \geq 20(t + u) - 250,$$
which implies the claim.

Analogously, $y \geq -12.5$.

From $x > 0 > y$, $x \leq 12.5$, $y \geq -12.5$, and $x - y \leq 20$, it follows that:
$$a_{10} \cdot a_{20} + a_{30} \cdot a_{40} \leq \frac{x^2 + y^2}{4} \leq 6.25^2 + 3.75^2.$$

This is attainable when $a_{10} = a_{20} = 6.25$ and $a_{30} = a_{40} = -3.75$. Therefore, the greatest possible value of $a_{10} \cdot a_{20} + a_{30} \cdot a_{40}$ is:
$$\boxed{6.25^2 + 3.75^2}.$$