Given that $\tan A=\frac{12}{5}$, $\cos B=-\frac{3}{5}$ and that $A$ and $B$ are in the same quadrant, find the value of $\cos (A-B)$.

$ \textbf{(A) }-\frac{63}{65}\qquad\textbf{(B) }-\frac{64}{65}\qquad\textbf{(C) }\frac{63}{65}\qquad\textbf{(D) }\frac{64}{65}\qquad\textbf{(E) }\frac{65}{63} $
1. **Determine the values of $\sin A$ and $\cos A$:**
   Given $\tan A = \frac{12}{5}$, we can use the identity $\tan A = \frac{\sin A}{\cos A}$.
   Let $\sin A = 12k$ and $\cos A = 5k$ for some $k$. Using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:
   \[
   (12k)^2 + (5k)^2 = 1 \implies 144k^2 + 25k^2 = 1 \implies 169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \pm \frac{1}{13}
   \]
   Since $A$ is in the third quadrant, both $\sin A$ and $\cos A$ are negative:
   \[
   \sin A = -12 \cdot \frac{1}{13} = -\frac{12}{13}, \quad \cos A = -5 \cdot \frac{1}{13} = -\frac{5}{13}
   \]

2. **Determine the value of $\sin B$:**
   Given $\cos B = -\frac{3}{5}$ and knowing $B$ is in the third quadrant, $\sin B$ must be negative. Using the Pythagorean identity:
   \[
   \sin^2 B + \cos^2 B = 1 \implies \sin^2 B + \left(-\frac{3}{5}\right)^2 = 1 \implies \sin^2 B + \frac{9}{25} = 1 \implies \sin^2 B = 1 - \frac{9}{25} = \frac{16}{25} \implies \sin B = -\frac{4}{5}
   \]

3. **Use the angle subtraction formula for cosine:**
   The formula for $\cos (A - B)$ is:
   \[
   \cos (A - B) = \cos A \cos B + \sin A \sin B
   \]
   Substituting the known values:
   \[
   \cos (A - B) = \left(-\frac{5}{13}\right) \left(-\frac{3}{5}\right) + \left(-\frac{12}{13}\right) \left(-\frac{4}{5}\right)
   \]
   Simplify each term:
   \[
   \cos (A - B) = \frac{15}{65} + \frac{48}{65} = \frac{63}{65}
   \]

The final answer is $\boxed{\frac{63}{65}}$