已知 $\sin A = -\frac{4}{5}$，$\tan B = -\frac{5}{12}$ 且 $\cos A > 0$，其中 $A$ 和 $B$ 处于不同的象限，在不使用计算器的情况下，求下列各式的值：

1. Determine the quadrant of $A$ and find $\cos A$

Given $\sin A = -\frac{4}{5}$ and $\cos A > 0$, $A$ lies in the 4th quadrant.
Using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:
$\cos^2 A = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$
Since $\cos A > 0$, $\cos A = \frac{3}{5}$.

2. Evaluate $\cot A$

$\cot A = \frac{\cos A}{\sin A} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$

1. Determine the quadrant of $A$ and find $\cos A$

Since $\sin A = -\frac{4}{5}$ and $\cos A > 0$, $A$ is in the 4th quadrant.
Using Pythagoras' theorem or the identity $\sin^2 A + \cos^2 A = 1$, we have $\cos A = \frac{3}{5}$.

2. Determine the quadrant of $B$ and find $\sin B$ and $\cos B$

Since $\tan B = -\frac{5}{12}$ (negative), $B$ is in either the 2nd or 4th quadrant.
Given that $A$ and $B$ are in different quadrants, $B$ must be in the 2nd quadrant.
Thus, $\sin B = \frac{5}{13}$ and $\cos B = -\frac{12}{13}$.

3. Evaluate $\cos(A + B)$ using the addition formula

$\cos(A + B) = \cos A \cos B - \sin A \sin B$
$= \left(\frac{3}{5}\right)\left(-\frac{12}{13}\right) - \left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)$
$= -\frac{36}{65} - \left(-\frac{20}{65}\right)$
$= -\frac{16}{65}$

1. Determine the quadrant of $B$ and find $\cos B$.

Given $\sin A = -\frac{4}{5}$ and $\cos A > 0$, $A$ is in the 4th quadrant.
Since $\tan B = -\frac{5}{12} < 0$ and $B$ is in a different quadrant from $A$, $B$ must be in the 2nd quadrant.
In the 2nd quadrant, $\cos B$ is negative.
Using a right-angled triangle with opposite $5$ and adjacent $12$, the hypotenuse is $\sqrt{5^2 + 12^2} = 13$.
Thus, $\cos B = -\frac{12}{13}$.

2. Use the double-angle identity to find $\sin^2\left(\frac{B}{2}\right)$.

Using the identity $\cos B = 1 - 2\sin^2\left(\frac{B}{2}\right)$:
$-\frac{12}{13} = 1 - 2\sin^2\left(\frac{B}{2}\right)$
$2\sin^2\left(\frac{B}{2}\right) = 1 + \frac{12}{13} = \frac{25}{13}$
$\sin^2\left(\frac{B}{2}\right) = \frac{25}{26}$

3. Determine the sign of $\sin\left(\frac{B}{2}\right)$ and evaluate the final answer.

Since $B$ is in the 2nd quadrant, $90^\circ < B < 180^\circ$.
Therefore, $45^\circ < \frac{B}{2} < 90^\circ$, which means $\frac{B}{2}$ is in the 1st quadrant, so $\sin\left(\frac{B}{2}\right) > 0$.
$\sin\left(\frac{B}{2}\right) = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}} = \frac{5\sqrt{26}}{26}$