3. Given that for any $m \in\left[\frac{1}{2}, 3\right]$, we have $x^{2}+m x+4>2 m+4 x$.
Then the range of values for $x$ is
$$
\begin{array}{l}
x>2 \text { or } x<2 m+4 x \\
\Leftrightarrow m(x-2)+(x-2)^{2}>0 . \\
\text { Let } g(m)=m(x-2)+(x-2)^{2}, \text { where, }
\end{array}
$$
$$
m \in\left[\frac{1}{2}, 3\right] \text {. }
$$

The problem is converted to $g(m)$ being always greater than 0 for $m \in\left[\frac{1}{2}, 3\right]$.
$$
\text { Then }\left\{\begin{array}{l}
g\left(\frac{1}{2}\right)>0, \\
g(3)>0 .
\end{array} \text { Solving, we get } x>2 \text { or } x<-1\right. \text {. }
$$