4. Given 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
4. $(0,1)$.

According to the problem, we know that $f^{\prime}(x)=0$ has two distinct zeros in the interval $(0,2)$, which means $x^{2}+b x+c=0$ has two distinct real roots in the interval $(0,2)$, let's denote them as $x_{1} 、 x_{2}$ $\left(x_{1} 、 x_{2} \in(0,2), x_{1} \neq x_{2}\right)$.
Then $x_{1}+x_{2}=-b, x_{1} x_{2}=c$
$$
\begin{aligned}
\Rightarrow & c^{2}+2 b c+4 c=c(c+2 b+4) \\
& =x_{1} x_{2}\left(x_{1} x_{2}-2 x_{1}-2 x_{2}+4\right) \\
& =x_{1} x_{2}\left(2-x_{1}\right)\left(2-x_{2}\right) \\
& =\left(x_{1}\left(2-x_{1}\right)\right)\left(x_{2}\left(2-x_{2}\right)\right) \\
& <\left(\frac{x_{1}+\left(2-x_{1}\right)}{2}\right)^{2}\left(\frac{x_{2}+\left(2-x_{2}\right)}{2}\right)^{2} \\
& =1 .
\end{aligned}
$$

Therefore, the range of $c^{2}+2 b c+4 c$ is $(0,1)$.