Let the coordinate system be +x East and +y North. The velocity of the water relative to the ground is $\vec{v}_{wg} = v_w \hat{i}$. The velocity of the boat relative to the water is $\vec{v}_{bw} = -v_b \sin(\theta) \hat{i} + v_b \cos(\theta) \hat{j}$.

The boat's velocity relative to the ground, $\vec{v}_{bg}$, is the vector sum:

$$\vec{v}_{bg} = \vec{v}_{bw} + \vec{v}_{wg}$$ $$\vec{v}_{bg} = (v_w - v_b \sin(\theta)) \hat{i} + (v_b \cos(\theta)) \hat{j}$$

[Q1] The magnitude is found using the Pythagorean theorem:

$$v_{bg} = \sqrt{(v_w - v_b \sin(\theta))^2 + (v_b \cos(\theta))^2}$$ $$v_{bg} = \sqrt{v_w^2 - 2v_w v_b \sin(\theta) + v_b^2 \sin^2(\theta) + v_b^2 \cos^2(\theta)}$$

Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$:

$$v_{bg} = \sqrt{v_b^2 + v_w^2 - 2v_w v_b \sin(\theta)}$$

[Q2] The direction is given by the angle $\alpha$ from the North (+y) axis.

$$\tan(\alpha) = \frac{|v_{bgx}|}{v_{bgy}} = \frac{|v_w - v_b \sin(\theta)|}{v_b \cos(\theta)}$$

The direction is East of North if $v_w > v_b \sin(\theta)$, and West of North if $v_w < v_b \sin(\theta)$.

[Q3] The time to cross the river depends only on the river width $W$ and the velocity component perpendicular to the banks, $v_{bgy}$.

$$t = \frac{W}{v_{bgy}} = \frac{W}{v_b \cos(\theta)}$$