In the first part $\displaystyle CD = 250\cos 75$
Therefore the distance $\displaystyle AD = \sqrt{ 350^2+ (300+250\cos 75)^2} $
Speed = Distance $\displaystyle \div $ Time = $\displaystyle \frac{\sqrt{ 350^2+ (300+250\cos 75)^2} }{12\times 60}$ metres per second
The steps are the following. You need to find Dickie's walking speed.
$\displaystyle \mbox{walking speed} = \frac{\mbox{distance travelled}}{\mbox{time travelled}}$.
You know "time travelled", so you only need to find "distance travelled."
You can apply Pythagoras' theorem to find the distance from D to A if you know AB and BD.
AB is given (350 meters), and BD = CD + 300 meters, so you need to find CD, which you can derive from the fact that CE as well as the angle at D are given:
$\displaystyle \tan(75) = \frac{EC}{CD}$,
so $\displaystyle CD\cdot\tan(75) = EC$, and $\displaystyle CD=\frac{EC}{\tan(75)}=EC\cdot\cot(75)=250\cdot \cot(75)$.
Now $\displaystyle AD=\sqrt{AD^2+BD^2}=\sqrt{350^2+(300+250\cdot\cot( 75))^2}$,
so the speed is $\displaystyle \sqrt{350^2+(300+250\cdot\cot(75))^2}/12$ meters per minute. (Convert to whatever unit required.)
Try to understand the steps above, and then solve problem (b) on your own.