If $$\theta$$ lies in the first quadrant and $$\cos^2 \theta - \sin^2 \theta = \frac{1}{2}$$, then the value of $$\tan^2 2\theta + \sin^2 3\theta$$ is:

$$\cos^2 \theta - \sin^2 \theta = \frac{1}{2}$$
$$\cos^2 \theta - (1 - \cos^2 \theta) = \frac{1}{2}$$
$$2\cos^2 \theta - 1 = \frac{1}{2}$$
$$2\cos^2 \theta = \frac{3}{2}$$
$$\cos^2 \theta = \frac{3}{4}$$
$$\cos \theta = \frac{\sqrt{3}}{2}$$
$$\theta = 30\degree$$
Now,
$$\tan^2 2\theta + \sin^2 3\theta$$
= $$\tan^2 2\times30\degree + \sin^2 3\times30\degree$$
= $$\tan^2 60\degree + \sin^2 90\degree$$
= 3 + 1 = 4