1. The problem is to verify the exact values of trigonometric functions for angles $-\frac{\pi}{12}$, $\frac{\pi}{12}$, and $\frac{5\pi}{12}$.\n\n2. We use the angle sum and difference identities for sine and tangent: \n- $\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b$\n- $\tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$\n\n3. Calculate $\sin(-\frac{\pi}{12})$: \nExpress $-\frac{\pi}{12}$ as $-\left(\frac{\pi}{4} - \frac{\pi}{6}\right)$. Using $\sin(-x) = -\sin x$, \n$$\sin\left(-\frac{\pi}{12}\right) = -\sin\left(\frac{\pi}{12}\right) = -\sin\left(\frac{\pi}{4} - \frac{\pi}{6}\right)$$\nApply sine difference formula: \n$$= -\left(\sin\frac{\pi}{4} \cos\frac{\pi}{6} - \cos\frac{\pi}{4} \sin\frac{\pi}{6}\right)$$\nSubstitute known values: $\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{6} = \frac{1}{2}$\n$$= -\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = -\left(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}\right) = -\frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$$\nThis matches the given value $(\sqrt{2} + \sqrt{6})/4$ only if signs are carefully checked; the correct value is $\frac{\sqrt{2} - \sqrt{6}}{4}$.\n\n4. Calculate $\tan(\frac{\pi}{12})$: \nExpress $\frac{\pi}{12}$ as $\frac{\pi}{4} - \frac{\pi}{6}$. Using tangent difference formula: \n$$\tan\left(\frac{\pi}{12}\right) = \frac{\tan\frac{\pi}{4} - \tan\frac{\pi}{6}}{1 + \tan\frac{\pi}{4} \tan\frac{\pi}{6}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$\nRationalize denominator: \n$$= \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$\nThis matches the given value.\n\n5. Calculate $\tan(\frac{5\pi}{12})$: \nExpress $\frac{5\pi}{12}$ as $\frac{\pi}{4} + \frac{\pi}{3}$. Using tangent sum formula: \n$$\tan\left(\frac{5\pi}{12}\right) = \frac{\tan\frac{\pi}{4} + \tan\frac{\pi}{3}}{1 - \tan\frac{\pi}{4} \tan\frac{\pi}{3}} = \frac{1 + \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}}$$\nRationalize denominator: \n$$= \frac{(1 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{(1 + \sqrt{3})^2}{1 - 3} = \frac{1 + 2\sqrt{3} + 3}{-2} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$$\nThe given value is $2 + \sqrt{3}$, so the correct value is $-2 - \sqrt{3}$.\n\n6. Calculate $\sin(\frac{5\pi}{12})$: \nExpress $\frac{5\pi}{12}$ as $\frac{\pi}{4} + \frac{\pi}{3}$. Using sine sum formula: \n$$\sin\left(\frac{5\pi}{12}\right) = \sin\frac{\pi}{4} \cos\frac{\pi}{3} + \cos\frac{\pi}{4} \sin\frac{\pi}{3} = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{2} + \sqrt{6}}{4}$$\nThis matches the given value.\n\nFinal answers:\n- $\sin\left(-\frac{\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$\n- $\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$\n- $\tan\left(\frac{5\pi}{12}\right) = -2 - \sqrt{3}$\n- $\sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{2} + \sqrt{6}}{4}$