1. The problem asks to find the values of $\sin 135^\circ$ and $\tan 15^\circ$.\n\n2. Recall the formulas and identities:\n- $\sin(180^\circ - \theta) = \sin \theta$\n- $\tan(45^\circ - \theta) = \frac{\tan 45^\circ - \tan \theta}{1 + \tan 45^\circ \tan \theta}$\n\n3. Calculate $\sin 135^\circ$:\nSince $135^\circ = 180^\circ - 45^\circ$, we have $\sin 135^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$.\n\n4. Calculate $\tan 15^\circ$ using the difference formula with $15^\circ = 45^\circ - 30^\circ$:\n$$\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$$\nWe know $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Substitute:\n$$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$\nMultiply numerator and denominator by $\sqrt{3} - 1$ to rationalize denominator:\n$$\tan 15^\circ = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$\n\nFinal answers:\n$$\sin 135^\circ = \frac{\sqrt{2}}{2}$$\n$$\tan 15^\circ = 2 - \sqrt{3}$$