1. Given the function $f(x)=x|1-x|(x \in \mathbf{R})$. Then the solution set of the inequality $f(x)>\frac{1}{4}$ is $\qquad$ .
$$
-1 \cdot\left(\frac{1+\sqrt{2}}{2},+\infty\right) \text {. }
$$

Notice that
$$
f(x)=x|1-x|=\left\{\begin{array}{ll}
-x^{2}+x, & x \leqslant 1 ; \\
x^{2}-x, & x>1 .
\end{array}\right.
$$

As shown in Figure 4, it is known that there exists $x_{0}$ in the interval $(1,+\infty)$ such that $f\left(x_{0}\right)=\frac{1}{4}$.
Let $x^{2}-x=\frac{1}{4}$.
Solving for $x$ gives $x=\frac{1 \pm \sqrt{2}}{2}$.
$$
\text { Also, } x \in(1,+\infty) \text {, }
$$

then $x_{0}=\frac{1+\sqrt{2}}{2}$.
Therefore, the solution set of the inequality is $\left(\frac{1+\sqrt{2}}{2},+\infty\right)$.