#### Answer

The ground speed is 380.4 mph
The actual bearing of the plane is $64^{\circ}$

#### Work Step by Step

Let $a = 355~mph$
Let $b = 28.5~mph$
Let angle $\theta$ be the angle between these two vectors. Then $\theta = 90^{\circ}-62^{\circ} = 28^{\circ}$
Let $c$ be the resultant of these two vectors. Note that the magnitude of $c$ is the ground speed.
We can use the parallelogram rule to find $c$:
$c = \sqrt{a^2+b^2+2ab~cos~\theta}$
$c = \sqrt{(355~mph)^2+(28.5~mph)^2+(2)(355~mph)(28.5~mph)~cos~28^{\circ}}$
$c = \sqrt{144703.7~mph^2}$
$c = 380.4~mph$
The ground speed is 380.4 mph
We can use the law of sines to find the angle B between the ground speed vector and the airspeed vector:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(28.5)~sin~152^{\circ}}{380.4})$
$B = arcsin(0.03517)$
$B = 2.0^{\circ}$
The actual bearing of the plane is $62^{\circ}+2.0^{\circ} = 64^{\circ}$