\[
\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)
\]
Calculate arguments:
\[
\arg(z_1) = \tan\left(\frac{4}{3}\right), \quad \arg(z_2) = \tan\left(\frac{-2}{1}\right) = -\tan(2)
\]
Therefore,
\[
\arg\left(\frac{z_1}{z_2}\right) = \tan\left(\frac{4}{3}\right) - (-\tan(2)) = \tan\left(\frac{4}{3}\right) + \tan(2)
\]
Using addition formula:
\[
\tan x + \tan y = \tan\left(\frac{x + y}{1 - xy}\right) \quad \text{(adjust for quadrant if necessary)}
\]
Substitute:
\[
x = \frac{4}{3}, \quad y = 2
\]
Calculate numerator and denominator:
\[
x + y = \frac{4}{3} + 2 = \frac{10}{3}
\]
\[
1 - xy = 1 - \frac{4}{3} \times 2 = 1 - \frac{8}{3} = -\frac{5}{3}
\]
So,
\[
\arg\left(\frac{z_1}{z_2}\right) = \tan\left(\frac{\frac{10}{3}}{-\frac{5}{3}}\right) = \tan(-2)
\]
Since denominator is negative, add \(\pi\):
\[
\arg = \tan(2) + \pi = \tan\left(\frac{10}{7}\right)
\]
Hence option (A) is correct.