Consider the following transformation:
$$\left( \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \right) - \left( \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a} \right) = \frac{a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c}{abc}$$
Now we transform only numerator:
$$\begin{aligned} a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c &= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) = \\ &= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) = (b-a)(a-c)(b^2 + ab - c^2 - ca) = \\ &= (b-a)(a-c)(b-c)(a+b+c). \end{aligned}$$
Therefore the condition $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=\frac{a^2}{c}+\frac{c^2}{b}+\frac{b^2}{a}$ for non-zero numbers $a, b, c$ is equivalent to condition: $(b-a)(a-c)(b-c)(a+b+c)=0$.

a) Similarly we conduct following transformation:
$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) - \left(\frac{a}{c} + \frac{c}{b} + \frac{b}{a}\right) = \frac{a^2 c - a^2 b + b^2 a - a^2 b + c^2 b - b^2 c}{abc} = \\ \frac{(b-a)(c-a)(c-b)}{abc} = 0.$$
Now, as a counter example let's choose a triple $a=1, b=2, c=-3$. It satisfies the first equation because $a+b+c=0$. However, the second equation is not correct:
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{1}{2} - \frac{2}{3} + \frac{3}{1} = \frac{3-4+18}{6} = \frac{17}{6}, \quad \frac{a}{c} + \frac{c}{b} + \frac{b}{a} = -\frac{1}{3} + \frac{3}{2} + \frac{2}{1} = \frac{-2+9+12}{6} = \frac{19}{6}.$$

6) From transformations above we conclude that the required triples can be two types
Type 1. Integers $a, b, c$ are pairwise distinct.
Type 2. If at least two integers are equal, for example $a=b \neq c$, then second equation is correct, while second is correct only if $c = -(a+b)$.
Now we count the number of such triples. The amount of pairwise different non-zero triples is $4040 \cdot 4039 \cdot 4038$. For the Type 2, if $a=b \neq c$, then there are $2020$ numbers to choose $a$ and then $b$ and $c$ are defined. Therefore, there are $3 \cdot 2020$ such triples.