1. Given the equation $4 x|x|-a x+1=0$ has exactly three solutions. Then the range of values for $a$ is
-1.4 .
Let the range of values for $a$ be $S$. Clearly, $0 \notin S$.
From $4 x|x|-a x+1=0$
$$
\Leftrightarrow a=\frac{4 x|x|+1}{x}=\left\{\begin{array}{l}
4 x+\frac{1}{x}, x>0 ; \\
\frac{1}{x}-4 x, x<0
\end{array}\right.
$$