Solution:

The maximum possible value is $18$, and it is achieved by all and only the triples that, in addition to the condition $a^{2}+b^{2}+c^{2}=6$, also satisfy $a+b+c=0$ (for example the triple with $a=b=1$ and $c=-2$).
To prove this it suffices to observe that
$$\begin{aligned} (a-b)^{2}+(b-c)^{2}+(c-a)^{2} & =2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a) \\ & =3\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\ & =18-(a+b+c)^{2} \\ & \leq 18 \end{aligned}$$
and that in the last step equality holds if and only if $a+b+c=0$.

b.

The maximum possible value is $108$, and it is achieved by all and only the triples in which the three variables are, in some order, $0$ and $\pm \sqrt{3}$.
We start by observing that, up to permutations, we can always assume that the three numbers satisfy the relation $a \leq b \leq c$. Recall also that
$$xy=\frac{(x+y)^{2}-(x-y)^{2}}{4} \leq \frac{(x+y)^{2}}{4}$$
for every pair of real numbers $x$ and $y$ (this is essentially the classical inequality between the geometric mean and the arithmetic mean), with equality if and only if $x=y$. Applying this inequality with $x=b-a$ and $y=c-b$ we then get that
$$(b-a) \cdot(c-b) \leq \frac{(c-a)^{2}}{4}$$
from which, squaring both sides (which does not change the direction of the inequalities since $b-a$ and $c-b$ are greater than or equal to $0$), we deduce that
$$(a-b)^{2} \cdot(b-c)^{2} \cdot(c-a)^{2} \leq \frac{(c-a)^{6}}{16} .$$
Finally we observe that
$$(c-a)^{2}=c^{2}+a^{2}-2 a c=2\left(a^{2}+c^{2}\right)-(a+c)^{2} \leq 2\left(a^{2}+b^{2}+c^{2}\right)=12,$$
from which we conclude that
$$(a-b)^{2} \cdot(b-c)^{2} \cdot(c-a)^{2} \leq \frac{12^{3}}{16}=108$$
To have equality, equality must hold in (2), which happens if and only if $b=0$ and $c=-a$, a condition that also guarantees that $x=y$ and hence equality in (1).

Alternative solution to question (b)

Let us set
$$x=a-b, \quad y=b-c, \quad z=c-a \text{.}$$
Up to permutations, we can arrange for $x \geq 0$ and $y \geq 0$ (it suffices that $c \leq b \leq a$). From question (a) we know that $x, y, z$ satisfy the inequality
$$x^{2}+y^{2}+z^{2} \leq 18 \text{,}$$
as well as of course the equality $x+y+z=0$, that is, $z=-(x+y)$. What we need to do is maximize the product $x^{2} y^{2} z^{2}$. Setting $s=x+y$ and $p=x y$, we observe that $s \geq 0$ and $p \geq 0$, and moreover
$$x^{2}+y^{2}+z^{2}=x^{2}+y^{2}+(x+y)^{2}=(x+y)^{2}-2 x y+(x+y)^{2}=2 s^{2}-2 p,$$
which allows us to rewrite condition (4) as $2 s^{2}-2 p \leq 18$, that is, $s^{2} \leq 9+p$. Finally, by the inequality between the arithmetic and geometric means, already mentioned in the first solution, we know that
$$p=x y \leq \frac{s^{2}}{4} .$$
Putting these facts together we deduce that
$$s^{2} \leq 9+p \leq 9+\frac{s^{2}}{4}$$
that is, $4 s^{2} \leq 36+s^{2}$, from which we conclude that $s^{2} \leq 12$, and hence
$$x^{2} y^{2} z^{2}=p^{2} s^{2} \leq\left(\frac{s^{2}}{4}\right)^{2} s^{2}=\frac{s^{6}}{16} \leq \frac{12^{3}}{16}=108 .$$
To have equality we need in particular that $s^{2}=12$ and $x=y$ (a necessary and sufficient condition for equality between the arithmetic and geometric means), from which $x^{2}=y^{2}=3$, that is (recalling that $x$ and $y$ are non-negative) $x=y=\sqrt{3}$. Returning then to equalities (3) we obtain that
$$a=b+\sqrt{3}, \quad \text{and} \quad c=b-\sqrt{3},$$
and in particular
$$6=a^{2}+b^{2}+c^{2}=(b+\sqrt{3})^{2}+b^{2}+(b-\sqrt{3})^{2}=3 b^{2}+6,$$
from which we conclude that $b=0$ and consequently $a=\sqrt{3}$ and $c=-\sqrt{3}$.