Answer: 2.

Solution. Suppose the leading coefficient of the polynomial is positive. Note that $(f(x))^{3}-4 f(x)=f(x) \cdot(f(x)-2) \cdot(f(x)+2)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-2$, and fewer roots than the equation $f(x)=2$. It is also clear that no two equations have common roots. Then the equation $f(x)=0$ has exactly one root. Therefore, the equation $f(x)=1$ has exactly two roots, and the equation $f(x)=-1$ has no roots. Thus, the equation $(f(x))^{2}-1=(f(x)+1)(f(x)-1)=0$ has two roots. Similarly, the case when the leading coefficient of the polynomial is negative is considered.
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-4 f(x)=0$ has exactly three solutions. How many solutions does the equation $(f(x))^{2}=1$ have?