2. Let $f(x)$ be a continuous even function, and for $x>0$, $f(x)$ is a strictly increasing function. Then the range of $x$ that satisfies
$$
f(x)<f\left(\frac{x+3}{x+4}\right)
$$

is
2. $\left(\frac{-5-\sqrt{13}}{2},-4\right) \cup\left(-4, \frac{-3-\sqrt{21}}{2}\right) \cup$
$$
\left(\frac{-5+\sqrt{13}}{2}, \frac{-3+\sqrt{21}}{2}\right) \text {. }
$$

Discuss the following three cases.
(1) $\left\{\begin{array}{l}x \geqslant 0, \\ \frac{x+3}{x+4}>0, \\ x0 \\ -x<\frac{x+3}{x+4}\end{array}\right.$
$\Rightarrow \frac{-5-\sqrt{13}}{2}<x<-4$ or $\frac{-5+\sqrt{13}}{2}<x<0$;
(3) $\left\{\begin{array}{l}x<0, \\ \frac{x+3}{x+4}<0, \\ -x<-\frac{x+3}{x+4}\end{array} \Rightarrow-4<x<\frac{-3-\sqrt{21}}{2}\right.$.
In summary,
$$
\begin{array}{l}
x \in\left(\frac{-5-\sqrt{13}}{2},-4\right) \cup\left(-4, \frac{-3-\sqrt{21}}{2}\right) \cup \\
\left(\frac{-5+\sqrt{13}}{2}, \frac{-3+\sqrt{21}}{2}\right) .
\end{array}
$$