$f^{\prime}(x)=x^{2}+b x+c$ has two distinct real roots in $(0,2)$ $\Rightarrow\left\{\begin{array}{l}c>0, \\ 4+2 b+c>0, \\ -\frac{b}{2} \in(0,2), \\ b^{2}-4 c>0\end{array}\right.$
$$
\Rightarrow c^{2}+2 b c+4 c<\frac{b^{4}}{16}+\frac{b^{3}}{2}+b^{2}=\left(\frac{b^{2}}{4}+b\right)^{2}=\left[\left(\frac{b}{2}+1\right)^{2}-1\right]^{2} \leqslant 1,
$$

Therefore, the range of $c^{2}+2 b c+4 c$ is $(0,1)$.
4. The function $f(x)=\frac{1}{3} x^{3}+\frac{1}{2} b x^{2}+c x+d$ has both a maximum and a minimum value in the interval $(0,2)$, then the range of $c^{2}+2 b c+4 c$ is $\qquad$ .