John is at O. He travels 40 km on a bearing of 210 degrees to town T. Attah travels from T on a bearing of 300 degrees for 35 km to his house. Calculate the coordinates of Attah's house.

Let John's initial position be the origin $O(0,0)$. John travels 40 km on a bearing of $210^\circ$.
To convert a bearing to a standard angle $\theta$ (measured counter-clockwise from the positive x-axis), we use $\theta = 90^\circ - \text{bearing}$.
For a bearing of $210^\circ$:

$$\theta_T = 90^\circ - 210^\circ = -120^\circ$$

The coordinates of T are $(x_T, y_T)$:

$$x_T = 40 \cos(-120^\circ) = 40 \left(-\frac{1}{2}\right) = -20$$ $$y_T = 40 \sin(-120^\circ) = 40 \left(-\frac{\sqrt{3}}{2}\right) = -20\sqrt{3}$$

So, the coordinates of town T are $(-20, -20\sqrt{3})$.

Attah moves 35 km from T on a bearing of $300^\circ$.
For a bearing of $300^\circ$:

$$\theta_A = 90^\circ - 300^\circ = -210^\circ$$

The displacement vector from T to Attah's final position is $(x_{disp}, y_{disp})$:

$$x_{disp} = 35 \cos(-210^\circ) = 35 \cos(150^\circ) = 35 \left(-\frac{\sqrt{3}}{2}\right) = -17.5\sqrt{3}$$ $$y_{disp} = 35 \sin(-210^\circ) = 35 \sin(150^\circ) = 35 \left(\frac{1}{2}\right) = 17.5$$

Attah's final position $(x_{Attah}, y_{Attah})$ is the sum of T's coordinates and Attah's displacement from T:

$$x_{Attah} = x_T + x_{disp} = -20 - 17.5\sqrt{3}$$ $$y_{Attah} = y_T + y_{disp} = -20\sqrt{3} + 17.5$$

Using $\sqrt{3} \approx 1.732$:

$$x_{Attah} = -20 - 17.5(1.732) = -20 - 30.31 = -50.31$$ $$y_{Attah} = -20(1.732) + 17.5 = -34.64 + 17.5 = -17.14$$

The position of Attah from John's initial position is approximately $(-50.31, -17.14)$ km.

The position of Attah from John's initial position is $\boxed{(-50.31, -17.14)}$.

The position vector of B is $\vec{OB} = \vec{OA} + \vec{AB}$.

$$\vec{OB} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 1+3 \\ 2+2 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$$

So, the coordinates of B are $\boxed{(4,4)}$.

The position vector of C is $\vec{OC} = \vec{OB} + \vec{BC}$.

$$\vec{OC} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4-1 \\ 4+3 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$$

So, the coordinates of C are $\boxed{(3,7)}$.

The position vector of D is $\vec{OD} = \vec{OC} + \vec{CD}$.

$$\vec{OD} = \begin{pmatrix} 3 \\ 7 \end{pmatrix} + \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 3+2 \\ 7-2 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}$$

So, the coordinates of D are $\boxed{(5,5)}$.