解: ∵ $\sin^2 A + \sin^2 B – \sin^2 C = \sin A \sin B$

∴ $a^2 + b^2 – c^2 = ab,∴ \cos C = \displaystyle \frac{a^2 + b^2 – c^2}{2ab} = \displaystyle \frac{1}{2}$

∵ $0 < C < \displaystyle \frac{\pi}{2},∴ C = \displaystyle \frac{\pi}{3} ∴ \sin C = \displaystyle \frac{\sqrt{3}}{2}$

∵ $\displaystyle \frac{\sin B\sin C}{3\sin A} = \displaystyle \frac{\cos A}{a} + \displaystyle \frac{\cos C}{c}$

∴ $\displaystyle \frac{\sqrt{3}}{2} \cdot \displaystyle \frac{b}{3a} = \displaystyle \frac{\sin C\cos A+\sin A\cos C}{\sin A\sin C}$

$= \displaystyle \frac{\sin(A+C)}{\sin A\sin C} = \displaystyle \frac{\sin B}{\sin A\sin C} = \displaystyle \frac{b}{ac}$

∴ $c = 2\sqrt{3}$

∵ $\displaystyle \frac{a}{\sin A} = \displaystyle \frac{b}{\sin B} = \displaystyle \frac{c}{\sin C} = 4$

∴ $a = 4\sin A, b = 4\sin B$

∴ $a + b = 4(\sin A + \sin B)$

$= 4\left[\sin A + \sin\left(\displaystyle \frac{2\pi}{3} – A\right)\right] = 4\sqrt{3}\sin\left(A + \displaystyle \frac{\pi}{6}\right)$

∵ $\begin{cases} 0 < A < \displaystyle \frac{\pi}{2} \\ 0 < B = \displaystyle \frac{2\pi}{3} - A < \displaystyle \frac{\pi}{2} \end{cases} \Rightarrow \displaystyle \frac{\pi}{6} < A < \displaystyle \frac{\pi}{2}$

∴ $A + \displaystyle \frac{\pi}{6} \in \left(\displaystyle \frac{\pi}{3}, \displaystyle \frac{2\pi}{3}\right)$

∴ $\sin\left(A + \displaystyle \frac{\pi}{6}\right) \in \left(\displaystyle \frac{\sqrt{3}}{2}, 1\right]$

∴ $a + b \in (6, 4\sqrt{3}]$

∴ $\displaystyle \frac{c^2}{a+b} \in [\sqrt{3}, 2)$