Calculate departure, latitude, and horizontal distance between two stations given their coordinates and instrument heights.

Departure is the change in Easting from Station A to Station B.

$$\Delta E = E_B - E_A$$ $$\Delta E = 315685.770 \text{ m} - 315420.550 \text{ m}$$ $$\Delta E = 265.220 \text{ m}$$

Latitude is the change in Northing from Station A to Station B.

$$\Delta N = N_B - N_A$$ $$\Delta N = 745495.610 \text{ m} - 745210.330 \text{ m}$$ $$\Delta N = 285.280 \text{ m}$$

The departure is $\boxed{\text{265.220 m}}$ and the latitude is $\boxed{\text{285.280 m}}$.

The horizontal distance is calculated using the Pythagorean theorem.

$$D = \sqrt{(\Delta E)^2 + (\Delta N)^2}$$ $$D = \sqrt{(265.220 \text{ m})^2 + (285.280 \text{ m})^2}$$ $$D = \sqrt{70341.6484 \text{ m}^2 + 81384.6784 \text{ m}^2}$$ $$D = \sqrt{151726.3268 \text{ m}^2}$$ $$D \approx 389.521 \text{ m}$$

The horizontal distance is $\boxed{\text{389.521 m}}$.

Since $\Delta E$ is positive and $\Delta N$ is positive, the azimuth is in the first quadrant.

$$\text{Azimuth} = \arctan\left(\frac{\Delta E}{\Delta N}\right)$$ $$\text{Azimuth} = \arctan\left(\frac{265.220}{285.280}\right)$$ $$\text{Azimuth} = \arctan(0.92968388)$$ $$\text{Azimuth} \approx 42.9197^\circ$$

Convert to degrees, minutes, and seconds:
$0.9197^\circ \times 60 = 55.182'$
$0.182' \times 60 = 10.92''$

$$\text{Azimuth} = 42^\circ 55' 11''$$

The whole circle horizontal azimuth angle is $\boxed{42^\circ 55' 11''}$.

The corrected change in elevation accounts for the instrument height at A and the target height at B.

$$\Delta H = (EL_B - EL_A) + (HT - HI)$$ $$\Delta H = (1881.960 \text{ m} - 1855.420 \text{ m}) + (1.800 \text{ m} - 1.550 \text{ m})$$ $$\Delta H = 26.540 \text{ m} + 0.250 \text{ m}$$ $$\Delta H = 26.790 \text{ m}$$

The vertical angle is calculated using the corrected change in elevation and the horizontal distance.

$$\theta_v = \arctan\left(\frac{\Delta H}{D}\right)$$ $$\theta_v = \arctan\left(\frac{26.790 \text{ m}}{389.5206 \text{ m}}\right)$$ $$\theta_v = \arctan(0.068779)$$ $$\theta_v \approx 3.9379^\circ$$

Convert to degrees, minutes, and seconds:
$0.9379^\circ \times 60 = 56.274'$
$0.274' \times 60 = 16.44''$

$$\theta_v = 3^\circ 56' 16''$$

The vertical angle is $\boxed{3^\circ 56' 16''}$.