For the diagram, please refer to Ross Millikan's answer. I am stealing his picture and labels. The main difference between our answers is that in the one below, we can proceed directly by calculator.
Let $\phi =\angle CDB$. By the Sine Law for $\triangle BDC$,
$$\frac{\sin\phi}{650}=\frac{\sin 12.5^\circ}{150}.$$
Now we can find $\phi$ to excellent accuracy with the calculator. I assume you know how to do that (calculate $\sin\phi$, press the $\sin^{-1}$ button). To $1$ decimal place, $\phi \approx 69.7^\circ$.
It follows that $\angle ABD$ is, to $1$ decimal place, $90^\circ-69.7\circ=20.3^\circ$. To find the angle the hill makes with the horizontal, subtract $12.5^\circ$. We get $7.8^\circ$. Convert the $0.8$ part to minutes if you wish.
Alternately, let $\theta=\angle ABD$. Then $\sin\phi=\cos\theta$, so the Sine Law gives directly $\frac{\cos\theta}{650}=\frac{\sin 12.5^\circ}{150}$.
The distances sailed are small enough that the curvature of the Earth makes no significant difference and we can use plane trigonometry.
Let $P$ be our start point, $Q$ where we change course, and $R$ the end point. First choose $P$. Then draw a North-South thin line through $P$ as a guide. To reach $Q$ we turned $40^\circ 10'$ clockwise from due South, and sailed $15$ km. Draw $Q$. Draw a thin North-South line through $Q$ as a guide. We sail $21$ km in a direction $28^\circ 20'$ counterclockwise from due North. Draw the point $R$ that we reach.
By properties of transversals to parallel lines (our two guide lines),
$$\angle Q=40^\circ 10'+28^\circ 20'=68^\circ30'.$$
We can now find the required distance $PR$ we are from our start position by using the Cosine Law. For
$$(PR)^2=15^2+21^2-2(15)(21)\cos 68^\circ30'.$$
(I get $PR\approx 20.86$, but my calculations are not to be trusted.)
Now for the direction. We will know everything once we know $\angle A$. For this, we could use the Cosine Law, but the Sine Law is easier. We have
$$\frac{\sin A}{21}=\frac{\sin 68.5^\circ}{PR}.$$
I get (but again don't trust me) that $A\approx 69.5^\circ$.
If we use the North-South guideline, our position at $R$, as viewed from $P$ is obtained by facing due South and turning clockwise through about $40^\circ10'+69^\circ 30'$. To express this in the notation that your problem was put, subtract from $180^\circ$. We get $70^\circ 20'$. So $R$ is North, $70^\circ 20'$ West from $P$.
Best Answer
We will assume that the Earth is flat, and that we are not too far North. Draw a picture. Let the share station be at $C$. For the first ship, go $42^\circ40'$ East from due North, and travel $155$ km. Call the resulting point $A$. So we face North from $C$, turn $42^\circ40'$ clockwise, and sail $155$ km.
For the second, go $45^\circ10'$ West from due North, and travel $165$ km. Call the resulting point $B$.
Then $\triangle ABC$ has $CA=155$, $CB=165$, and $\angle C=42^\circ40'+45^\circ10'=87^\circ 50'$.
By the Cosine Law, $$(AB)^2=155^2+165^2-2(155)(165)\cos C.$$ Calculate. It turns out that the distance is about $222$ km.
Remark: We can proceed more informally without the Cosine Law, and with a little crossing of the fingers. Note that $\angle C$ is almost a right angle. If we pretend it is a right angle, we can use the Pythagorean Theorem to estimate the distance. That gives $226$ km. Not very different from $222$.