Given z1 = 3 + 5i and z2 = 2 + 4i.

Given $z_1 = 3 + 5i$ and $z_2 = 2 + 4i$.
To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $z_2 = 2 + 4i$ is $2 - 4i$.

$$\frac{z_1}{z_2} = \frac{3 + 5i}{2 + 4i} \times \frac{2 - 4i}{2 - 4i}$$

Multiply the numerators:

$$(3 + 5i)(2 - 4i) = 3(2) + 3(-4i) + 5i(2) + 5i(-4i)$$ $$= 6 - 12i + 10i - 20i^2$$

Since $i^2 = -1$:

$$= 6 - 2i - 20(-1)$$ $$= 6 - 2i + 20 = 26 - 2i$$

Multiply the denominators:

$$(2 + 4i)(2 - 4i) = 2^2 - (4i)^2$$ $$= 4 - 16i^2$$

Since $i^2 = -1$:

$$= 4 - 16(-1) = 4 + 16 = 20$$

Now combine the numerator and denominator:

$$\frac{z_1}{z_2} = \frac{26 - 2i}{20} = \frac{26}{20} - \frac{2}{20}i$$ $$\frac{z_1}{z_2} = \frac{13}{10} - \frac{1}{10}i$$

Given $z_1 = 3 + 5i$. The modulus $|z_1|$ is calculated as $\sqrt{x^2 + y^2}$.

$$|z_1| = \sqrt{3^2 + 5^2}$$ $$= \sqrt{9 + 25}$$ $$= \sqrt{34}$$

Given $z_1 = 3 + 5i$. The argument $\arg(z_1)$ is $\theta$ such that $\tan \theta = \frac{y}{x}$. Since $x=3$ and $y=5$ are both positive, $z_1$ is in the first quadrant.

$$\tan \theta = \frac{5}{3}$$ $$\theta = \arctan\left(\frac{5}{3}\right)$$

In degrees:

$$\theta \approx 59.04^\circ$$

In radians:

$$\theta \approx 1.03 \text{ radians}$$

The results for the complex numbers are:

$$\boxed{\frac{z_1}{z_2} = \frac{13}{10} - \frac{1}{10}i}$$ $$\boxed{|z_1| = \sqrt{34}}$$ $$\boxed{\arg(z_1) = \arctan\left(\frac{5}{3}\right) \approx 59.04^\circ \text{ or } 1.03 \text{ radians}}$$

Use the logarithm property $\log_b M^p = p \log_b M$:

$$x \log_4 3 - 4 \log_4 3 = 0$$

Factor out $\log_4 3$:

$$(x - 4) \log_4 3 = 0$$

Since $\log_4 3 \neq 0$ (because $3 \neq 4^0 = 1$), we must have:

$$x - 4 = 0$$ $$x = 4$$

The solution to the logarithm equation is:

$$\boxed{x = 4}$$

Given $\sin A = \frac{4}{5}$ and $\cos B = \frac{12}{13}$. Angles $A$ and $B$ are both acute, meaning they are in the first quadrant.

First, find $\cos A$ and $\sin B$.
For angle $A$:
Since $A$ is acute, $\cos A > 0$.

$$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{4}{5}\right)^2}$$ $$= \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

For angle $B$:
Since $B$ is acute, $\sin B > 0$.

$$\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{12}{13}\right)^2}$$ $$= \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{169 - 144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$$

Now, calculate $\sin(A+B)$ using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

$$\sin(A+B) = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) + \left(\frac{3}{5}\right)\left(\frac{5}{13}\right)$$ $$= \frac{48}{65} + \frac{15}{65}$$ $$= \frac{48 + 15}{65} = \frac{63}{65}$$

Finally, calculate $\cos(A-B)$ using the formula $\cos(A-B) = \cos A \cos B + \sin A \sin B$:

$$\cos(A-B) = \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{5}{13}\right)$$ $$= \frac{36}{65} + \frac{20}{65}$$ $$= \frac{36 + 20}{65} = \frac{56}{65}$$

The results for the trigonometric expressions are:

$$\boxed{\sin(A+B) = \frac{63}{65}}$$ $$\boxed{\cos(A-B) = \frac{56}{65}}$$