1. **Problem 1: Find coordinates of P(A) on the unit circle where $A = -\frac{11\pi}{6}$.** 2. The unit circle allows us to express coordinates as $(\cos A, \sin A)$. Here $A = -\frac{11\pi}{6}$ which is coterminal to $\frac{\pi}{6}$ because adding $2\pi$ we get $-\frac{11\pi}{6} + 2\pi = \frac{\pi}{6}$. 3. At angle $\frac{\pi}{6}$, the coordinates are $(\cos \frac{\pi}{6}, \sin \frac{\pi}{6}) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. 4. Because $A$ is negative and lies in the fourth quadrant, coordinates reflect accordingly: $\cos$ remains positive and $\sin$ is negative. 5. Therefore, coordinates at $A = -\frac{11\pi}{6}$ are $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ which corresponds to **Option 4**. 6. **Problem 2: Given $\tan A = \frac{3}{5}$ and $A$ is in quadrant III, find $\sec A$.** 7. Since $\tan A = \frac{\sin A}{\cos A} = \frac{3}{5}$ and $A$ is in QIII where $\sin A < 0$ and $\cos A < 0$, let $\sin A = -3k$ and $\cos A = -5k$ for some $k > 0$. 8. Using the Pythagorean identity, $\sin^2 A + \cos^2 A = 1$, we get: $$(-3k)^2 + (-5k)^2 = 1 \implies 9k^2 + 25k^2 = 1 \implies 34k^2 = 1 \implies k = \frac{1}{\sqrt{34}}$$ 9. Then $\cos A = -5k = -\frac{5}{\sqrt{34}}$. 10. $\sec A = \frac{1}{\cos A} = \frac{1}{-5 / \sqrt{34}} = -\frac{\sqrt{34}}{5}$. **Final answers:** - Coordinates of point $P(A)$: $\boxed{\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)}$ (Option 4). - Value of $\sec A$: $\boxed{-\frac{\sqrt{34}}{5}}$ (Option 2).