I agree with your (a) answer and Rezha's (b) looks good.
For (c), moving everything up by 10 gives you 4 nonnegative numbers with sum 58, but summands over 20 definitely come up a lot. Here's the trick: Move everything down 10 instead, then multiply everything by $-1$:
\begin{align}
a + b + c + d = 18 & \quad \text{with } -10 \le a, b, c, d \le 10 \\
(a-10) + (b-10) + (c-10) + (d-10) = 18-40 \\
\alpha + \beta + \gamma + \delta = -22 & \quad \text{with } -20 \le \alpha, \beta, \gamma, \delta \le 0 \\
-\alpha - \beta -\gamma - \delta = 22 \\
w + x + y + z = 22 & \quad \text{with } 0 \le w, x, y, z \le 20
\end{align}
That is, move the variables down 10 and then negate them. The range of allowed values goes from $[-10,10]$ to $[-20,0]$ to $[0,20]$, so a maximum constraint is still in effect. But the sum went from 18 to $18-40 = -22$ to 22: we'll see that the maximum constraint won't matter much when it's so close to the sum.
Count solutions to $w + x + y + z = 22$ as in (a), I believe you get ${25 \choose 3} = 2300$. Those include some solutions with values over 20, but very few: You could have $(22,0,0,0)$ with 4 ways to assign the 22, and $(21,1,0,0)$ with 12 ways to assign the 21 and 1 (not just ${4 \choose 2}=6$ ways since, e.g., $(0,21,0,1)$ and $(0,1,0,21)$ are distinct solutions). Removing those 16 "bad" solutions leaves 2284 with allowed values.
PS: In the time it takes to think of that and work out the details, you could find a computer algebra system and look up the coefficient of $x^{18}$ in the expansion of
$$(x^{-10} + x^{-9} + \cdots + x^{-1} + x^0 + x^1 + \cdots + x^9 + x^{10})^4$$
which is, in fact, 2284.