Let $$S = \{\theta \in [0, 2\pi) : \tan(\pi\cos\theta) + \tan(\pi\sin\theta) = 0\}$$, then $$\sum_{\theta \in S} \sin^2\left(\theta + \frac{\pi}{4}\right)$$ is equal to

$$\tan(\pi \cos \theta) = -\tan(\pi \sin \theta) = \tan(-\pi \sin \theta)$$

The general solution for $$\tan A = \tan B$$ is $$A = n\pi + B, \quad \text{where } n \in \mathbb{Z}$$

$$\pi \cos \theta = n\pi - \pi \sin \theta, \quad n \in \mathbb{Z}$$

$$\cos \theta + \sin \theta = n \quad \text{--- (Equation 1)}$$

$$-\sqrt{1^2 + 1^2} \le \sin \theta + \cos \theta \le \sqrt{1^2 + 1^2}$$

$$-\sqrt{2} \le \sin \theta + \cos \theta \le \sqrt{2}$$

$$n = -1, \, 0, \, 1$$

Case 1:

$$\cos \theta + \sin \theta = 1$$

$$\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = 1$$

$$1 + \sin 2\theta = 1 \implies \sin 2\theta = 0$$

$$\theta = 0, \, \frac{\pi}{2}$$

Case 2:

$$\cos \theta + \sin \theta = -1$$

Similarly, $$\theta = \pi, \, \frac{\pi}{3\pi}{2}$$

Case 3:

$$\cos \theta + \sin \theta = 0 \implies \tan \theta = -1$$

$$\theta = \frac{3\pi}{4}, \, \frac{7\pi}{4}$$

$$S = \left\{ 0, \, \frac{\pi}{2}, \, \pi, \, \frac{3\pi}{2}, \, \frac{3\pi}{4}, \, \frac{7\pi}{4} \right\}$$

$$\sum_{\theta \in S} \sin^2\left(\theta + \frac{\pi}{4}\right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 0 + 0 = 2$$