Answer
(a) To return to his starting point, Harrison should travel at an angle of $29.6^{\circ}$ north of east for a distance of $6.34~km$
(b) The return trip will take 1268 seconds.
Work Step by Step
We can find the total west component of Harrison's trip:
$2.00~km+(5.00~km)~cos~53.0^{\circ}+(1.00~km)~cos~60.0^{\circ} = 5.51~km$
We can find the total south component of Harrison's trip:
$(5.00~km)~sin~53.0^{\circ}-(1.00~km)~sin~60.0^{\circ} = 3.13~km$
To return to the starting point, Harrison should travel 5.51 km east and 3.13 km north.
We can find the angle north of east that Harrison should travel:
$tan~\theta = \frac{3.13}{5.51}$
$\theta = tan^{-1}(\frac{3.13}{5.51})$
$\theta = 29.6^{\circ}$
We can find the distance $d$ that Harrison should travel:
$d = \sqrt{(5.51~km)^2+(3.13~km)^2} = 6.34~km$
To return to his starting point, Harrison should travel at an angle of $29.6^{\circ}$ north of east for a distance of $6.34~km$
(b) We can find the time of the return trip:
$t = \frac{d}{v} = \frac{6340~m}{5.00~m/s} = 1268~seconds$
The return trip will take 1268 seconds.
