1. Let real numbers $a, b, c, d$ satisfy $a \geqslant b, c \geqslant d$, and $|a|+2|b|+3|c|+4|d|=1$. Let $P=(a-b)(b-c)(c-d)$, find the minimum and maximum values of $P$.
$$
\begin{array}{l}
\text { Since }|P| \leqslant(|a|+|b|)(|b|+|c|)(|c|+|d|)=\frac{1}{2}(|a|+|b|)(|b|+|c|)(2|c|+2|d|) \\
\leqslant \frac{1}{2}\left(\frac{|a|+2|b|+3|c|+2|d|}{3}\right)^{3} \leqslant \frac{1}{2}\left(\frac{|a|+2|b|+3|c|+4|d|}{3}\right)^{3}=\frac{1}{2} \cdot \frac{1}{27}=\frac{1}{54},
\end{array}
$$

Equality holds when $a=-b=c, d=0 \Rightarrow(a, b, c, d)=\left(\frac{1}{6},-\frac{1}{6}, \frac{1}{6}, 0\right)$.
Thus, the minimum value of $P$ is $-\frac{1}{54}$.
Assume $b>c$ (otherwise $P<0), a-b=x, b-c=y, c-d=z$, then
$$
\begin{array}{l}
1=|a|+2|b|+3|c|+4|d| \geqslant(|a|+|d|)+2(|b|+|d|)+(|c|+|d|) \\
\geqslant|a-d|+2|b-d|+|c-d|=x+y+z+2(y+z)+z=x+3 y+4 z \\
\geqslant 3 \sqrt[3]{12 x y z} \Rightarrow P=x y z \leqslant \frac{1}{12 \cdot 27}=\frac{1}{324} \text {. } \\
\end{array}
$$

Equality holds when $x=\frac{1}{3}, y=\frac{1}{9}, z=\frac{1}{12}, c=0 \Rightarrow(a, b, c, d)=\left(\frac{4}{9}, \frac{1}{9}, 0,-\frac{1}{12}\right)$. Thus, the maximum value of $P$ is $\frac{1}{324}$.