$c$ is the length of the line $AB$, i.e. from $(b \cos\gamma,\ b \sin\gamma)$ to $(a,\ 0)$, which has horizontal component $a - b \cos\gamma$ and vertical component $0 - b \sin\gamma$.
So using Pythagoras, $c = \sqrt{(a - b \cos\gamma)^2 + (0 - b \sin\gamma)^2}$.
Then expand, and simplify using Pythagoras again in $\sin^2 \gamma + \cos^2 \gamma =1$.
The angular displacement between $A$ and $C$ is $a_2 - a_1$, therefore, the minimum (perpendicular )distance from the origin to line $AC$ is given by
$d_1 = r_1 \cos \dfrac{(a_2 - a_1)}{2} $
similarly the minimum (perpendicular) distance from the origin to $BD$ is given by
$d_2 = r_2 \cos \dfrac{(a_2 - a_1)}{2} $
Hence the distance between the two lines is
$d = d_2 - d_1 = (r_2 - r_1) \cos \dfrac{(a_2 - a_1)}{2} $
For the second part, point $B'$ is given by
$B' = r_2 (\cos \phi, \sin \phi ) = A + t (C - A) $
Since the distance of $B'$ from the origin is $r_2$, then $t$ must be the root of the following equation:
$\mathbf{A} \cdot \mathbf{A} + 2 t \mathbf{A} \cdot (\mathbf{C}-\mathbf{A}) + t^2 (\mathbf{C} - \mathbf{A}) \cdot (\mathbf{C} - \mathbf{A}) = r_2 ^2$
All the quantities can be computed, and the quadratic equation in $t$ can be solved using the quadratic formula resulting in two values of $t$ corresponding to point $B'$ and $D'$. The negative value of $t$ corresponds to $B'$ and the positive value corresponds to $D'$. To find the angles just use the $\text{ATAN2}$ function, so that
$\phi_{\text{B'}} = \text{ATAN2}( B'_x , B'_y)$
$\phi_{\text{D'}} = \text{ATAN2}( D'_x , D'_y)$
Best Answer
This is a wrong answer because bearings are measured clockwise from north, while angles used in trigonometric calculations are usually measured anticlockwise from the positive x-direction, which on a map is east. Your idea with finding the vectors and adding them is completely correct, you just need to measure the angles correctly. The correct angle for your first bearing, for example, should be $-210° + 90° = -120° = 240°$, and your second angle should be $-290° + 90° = -200° = 160°$. If the answers still don't agree, it may be that the question is using relative bearings, but I think that's unlikely.