A

Given that \( \cos \theta = -\frac{12}{13} \), we can find \( \sin \theta \) and \( \tan \theta \) using the Pythagorean identity:
\( \sin^2 \theta + \cos^2 \theta = 1 \).
Firstly, calculate \( \sin^2 \theta \):
\( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \).
Therefore, \( \sin \theta = -\sqrt{\frac{25}{169}} = -\frac{5}{13} \) (negative in the third quadrant).
Now, calculate \( \tan \theta \):
\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{5}{13}}{-\frac{12}{13}} = \frac{5}{12} \).
Thus, the values are:
\( \sin \theta = -\frac{5}{13}, \tan \theta = \frac{5}{12} \). Therefore, the correct answer is option A.