1. Since $A + B + C = \pi$, hence $\tan \left(\frac{A+B}{2}\right) = \frac{1}{\tan \left(\frac{C}{2}\right)}$.

From the identity $\tan \left(\frac{A+B}{2}\right) + \tan \left(\frac{C}{2}\right) = \frac{4\sqrt{3}}{3}$, we can derive that $\frac{1}{\tan \left(\frac{C}{2}\right)} + \tan \left(\frac{C}{2}\right) = \frac{4 \sqrt{3}}{3}$.

Solving the equation gives $\tan \left(\frac{C}{2}\right) = \sqrt{3}$ or $\tan \left(\frac{C}{2}\right) = \frac{\sqrt{3}}{3}$.

Since $C \in (0, \pi)$, we find $C = \frac{\pi}{3}$ or $C = \frac{2\pi}{3}$.

Given that $\sin B \cdot \sin C = \cos^2 \left(\frac{A}{2}\right)$, we have $\sin B \cdot \sin C = \frac{1}{2}(1 + \cos A) = \frac{1}{2}(1 - \cos (B + C))$.

Thus $2\sin B \cdot \sin C = 1 - (\cos B \cdot \cos C - \sin B \cdot \sin C)$, which gives $\cos (B - C) = 1$.

Hence, we conclude $B = C$.

So $B = C = \frac{\pi}{3}$. The solution $B = C = \frac{2\pi}{3}$ is rejected because it does not satisfy the condition in the problem.

Therefore, $A = \pi - B - C = \frac{\pi}{3}$.

Putting it all together, we have $$\boxed{A = B = C = \frac{\pi}{3}}.$$

2. Using the results from part (1) and the given conditions, we have $C = \frac{\pi}{3}$, hence $A + B = \frac{2\pi}{3}$.

Therefore, $\sin A + \sin B = \sin A + \sin \left( \frac{2\pi}{3} - A \right)$.

This simplifies to $\sin A + \frac{\sqrt{3}}{2}\cos A + \frac{1}{2}\sin A = \frac{3}{2}\sin A + \frac{\sqrt{3}}{2}\cos A$.

Next, we can write this as $\sqrt{3}\sin \left(A + \frac{\pi}{6}\right)$.

Since $A \in \left(0, \frac{2\pi}{3}\right)$, it follows that $\left(A + \frac{\pi}{6}\right) \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)$.

Hence, $\sin \left(A + \frac{\pi}{6}\right) \in \left(\frac{1}{2}, 1\right]$.

Thus, the range of values for $\sin A + \sin B$ is $$\boxed{\left(\frac{\sqrt{3}}{2}, \sqrt{3}\right]}.$$