For the max, we have $abcd\leq |abcd|\leq \left(\sqrt{\frac{|a|^2+|b|^2+|c|^2+|d|^2}{4}}\right)^4=\left(\frac{12}{4}\right)^2=9$. Se have equality if $|a|=|b|=|c|=|d|$ and so if we have the numbers $(+\sqrt{3},+\sqrt{3},-\sqrt{3},-\sqrt{3})$ For the min, we have to have one negative and three positive, or one positive and three negative. In both cases, it is equivalent to find the max of $abc(a+b+c)$ given $a,b,c\geq 0$ and $12=a^2+b^2+c^2+(a+b+c)^2=2(\sum_{cyc} a^2+\sum_{cyc} ab)$. By expansion and muirhead, we have $abc(a+b+c)\leq \frac{1}{12}\left(\sum a^2+\sum ab\right)^2=\frac{1}{12}6^2=3$, where we have equality if and only if $a=b=c=1$. Therefore we can achieve the minimum $-3$ with $(1,1,1,-3)$, $(-1,-1,-1,3)$ and permutations.

By AM-GM, $$a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)$$$$abc(a+b+c)\leq \frac{1}{3}\left(ab+bc+ca\right)^2\leq \frac{1}{12}\left( a^2+b^2+c^2+ab+bc+ca\right)^2$$$$= \frac{1}{48}\left( a^2+b^2+c^2+(a+b+c)^2\right)^2=3$$ Let $a,b,c,d$ be real numbers such that $a+b+c+d=0$. Prove that $$\sqrt{\frac{12}{7}}\leq\frac{a^2+b^2+c^2+d^2}{\sqrt{a^4+b^4+c^4+d^4}}\leq 2$$h