Step 1: Find $\frac{z_1}{z_2}$. 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: $$ \frac{z_1}{z_2} = \frac{26 - 2i}{20} = \frac{26}{20} - \frac{2}{20}i $$ $$ \frac{z_1}{z_2} = \frac{13}{10} - \frac{1}{10}i $$ Step 2: Determine the modulus of $z_1$. Given $z_1 = 3 + 5i$. The modulus $|z_1|$ is calculated as $\sqrt{x^2 + y^2}$. $$ |z_1| = \sqrt{3^2 + 5^2} $$ $$ = \sqrt{9 + 25} $$ $$ = \sqrt{34} $$ Step 3: Determine the argument of $z_1$. 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: $$ \boxed{\frac{z_1}{z_2} = \frac{13}{10} - \frac{1}{10}i} $$ $$ \boxed{|z_1| = \sqrt{34}} $$ $$ \boxed{\arg(z_1) = \arctan\left(\frac{5}{3}\right) \approx 59.04^\circ \text{ or } 1.03 \text{ radians}} $$ Step 4: Solve the logarithm equation $\log_4 3^x - 4 \log_4 3 = 0$. 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: $$ \boxed{x = 4} $$ Step 5: Find $\sin(A+B)$ and $\cos(A-B)$. 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: $$ \boxed{\sin(A+B) = \frac{63}{65}} $$ $$ \boxed{\cos(A-B) = \frac{56}{65}} $$