A and B are acute angles, such that $\cos A = \frac{3}{5}$ and $\tan B = \frac{12}{5}$.\n\nFind the exact values of

1. Find the exact value of $\tan A$.

Since $A$ is an acute angle, $\tan A > 0$.
From a right-angled triangle with adjacent side $3$ and hypotenuse $5$, the opposite side is $\sqrt{5^2 - 3^2} = 4$.
Thus, $\tan A = \frac{4}{3}$.

2. Calculate $\tan 2A$ using the double angle formula.

$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$
$\tan 2A = \frac{2\left(\frac{4}{3}\right)}{1 - \left(\frac{4}{3}\right)^2}$
$\tan 2A = \frac{\frac{8}{3}}{1 - \frac{16}{9}}$
$\tan 2A = \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{24}{7}$

1. Find the exact value of $\cos B$ using the given $\tan B$.

Given $\tan B = \frac{12}{5}$ and $B$ is an acute angle.
Using a right-angled triangle with opposite side $12$ and adjacent side $5$, the hypotenuse is $\sqrt{12^2 + 5^2} = 13$.
Therefore, $\cos B = \frac{5}{13}$.

2. Simplify the expression $\sin\left(B + \frac{3\pi}{2}\right)$ using the addition formula or allied angles.

Using the addition formula for sine:
$\sin\left(B + \frac{3\pi}{2}\right) = \sin B \cos\left(\frac{3\pi}{2}\right) + \cos B \sin\left(\frac{3\pi}{2}\right)$
Since $\cos\left(\frac{3\pi}{2}\right) = 0$ and $\sin\left(\frac{3\pi}{2}\right) = -1$,
$\sin\left(B + \frac{3\pi}{2}\right) = \sin B (0) + \cos B (-1) = -\cos B$

3. Substitute the value of $\cos B$ to find the final answer.

$\sin\left(B + \frac{3\pi}{2}\right) = -\cos B = -\frac{5}{13}$

1. Find the exact value of $\cos B$

Since $\tan B = \frac{12}{5}$ and $B$ is acute, we can form a right-angled triangle with opposite side $12$ and adjacent side $5$.
The hypotenuse is $\sqrt{12^2 + 5^2} = 13$.
Therefore, $\cos B = \frac{5}{13}$.

2. Apply the double angle identity to relate $\cos B$ and $\cos\frac{B}{2}$

Using the identity $\cos 2\theta = 2\cos^2\theta - 1$ with $\theta = \frac{B}{2}$:
$\cos B = 2\cos^2\frac{B}{2} - 1$
$\frac{5}{13} = 2\cos^2\frac{B}{2} - 1$
$2\cos^2\frac{B}{2} = 1 + \frac{5}{13} = \frac{18}{13}$

3. Solve for $\cos\frac{B}{2}$ and determine the correct sign

$\cos^2\frac{B}{2} = \frac{9}{13}$
$\cos\frac{B}{2} = \pm\frac{3}{\sqrt{13}}$
Since $B$ is an acute angle ($0 < B < 90^\circ$), $\frac{B}{2}$ is also acute ($0 < \frac{B}{2} < 45^\circ$), so $\cos\frac{B}{2} > 0$.
$\cos\frac{B}{2} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}$