2. Positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=\sqrt{14}, \\
x^{2}+z^{2}-\sqrt{2+\sqrt{3}} x z=7-4 \sqrt{3}, \\
y^{2}+z^{2}+\sqrt{2+\sqrt{3}} y z=7+4 \sqrt{3} .
\end{array}\right.
$$

Then $z=$
2. $\frac{\sqrt{3}+1}{\sqrt{7}}$.

According to the conditions of the system of equations, we can draw Figure 2.
Notice that
$$
\cos 15^{\circ}=\frac{1}{2} \sqrt{2+\sqrt{3}}, 7 \pm 4 \sqrt{3}=(2 \pm \sqrt{3})^{2},
$$

Therefore, equation (2) represents the third side $AC = 2 - \sqrt{3}$ of $\triangle ADC$ with sides $DA = x$ and $DC = z$, and the included angle $\angle ADC = 15^{\circ}$;

Equation (3) represents the third side $BC = 2 + \sqrt{3}$ of $\triangle BDC$ with sides $DB = y$, $DC = z$, and the included angle $\angle BDC = 165^{\circ}$.

Also, $(2-\sqrt{3})^{2}+(2+\sqrt{3})^{2}=(\sqrt{14})^{2}$, which means $\triangle ABC$ is a right triangle with $\angle C$ as the right angle. Hence, $\sin A=\frac{2+\sqrt{3}}{\sqrt{14}}$.
By the Law of Sines, $\frac{z}{\sin A}=\frac{2-\sqrt{3}}{\sin 15^{\circ}}$.
Solving for $z$ gives $z=\frac{2 \sqrt{2+\sqrt{3}}}{\sqrt{14}}=\frac{\sqrt{3}+1}{\sqrt{7}}$.