Answer
$$68.7\;\rm m/s, \;21.3^\circ \tag{with vertical}$$
Work Step by Step
We know that the velocity of the rain relative to the ground is given by
$$v_{RG}=v_{RC}+v_{CG}\tag 1$$
whereas $R\rightarrow$ rain, $G\rightarrow$ ground, $C\rightarrow$ car.
We chose the North-south line to be our $x$-aixs at which the positive $x$-axis is in the direction of north.
And we also chose up to be our positive $y$-direction.
So, when the car is moving north at a constant speed of 25 m/s, its velocity vector relative to the ground is
$$v_{CG}=(25)\hat i\tag 2$$
And the rain is making an angle of $38^\circ$ with vertical, which means its angle relative to our $+x$-axis is in the fourth quadrant, So,
$$\theta_{RC}=270^\circ +38^\circ=\bf 308^\circ$$
Thus, its velocity vector relative to the car is given by
$$v_{RC}=(v_{RC}\cos 308^\circ)\hat i+(v_{RC}\sin 308^\circ)\hat j\tag 3$$
From (1);
$$ v_{RC}=v_{RG}-v_{CG}=v_{RG}\cos\theta\;\hat i+v_{RG}\sin\theta \;\hat j- (25)\hat i$$
$$ v_{RC}=(v_{RG}\cos\theta-25)\hat i+(v_{RG}\sin\theta )\hat j$$
We know that the angle in the first case is 308$^\circ$.
Thus, $$\tan 308^\circ=\dfrac{v_{RG}\sin\theta }{v_{RG}\cos\theta-25}\tag 4$$
In the second case, when the car now moves south.
$$ v_{RC}=(v_{RG}\cos\theta+25)\hat i+(v_{RG}\sin\theta )\hat j$$
We know that the angle in the second case is 270$^\circ$.
Thus, $$\tan 270^\circ=\dfrac{v_{RG}\sin\theta }{v_{RG}\cos\theta+25}$$
hence,
$$v_{RG}\cos\theta=-25 \tag 5$$
Plug into (4);
$$\tan 308^\circ=\dfrac{v_{RG}\sin\theta }{-25-25} $$
Thus,
$$v_{RG}\sin\theta=64\tag 6$$
Thus, the rain velocity magnitude is given by (see (5), and (6))
$$|\vec v_{RG}|=\sqrt{25^2+64^2}=\color{red}{\bf 68.7}\;\rm m/s$$
and its direction;
$$\theta=\tan^{-1}\left[\dfrac{-64}{ 25}\right]=-68.7^\circ$$
which makes an angle of $21^\circ$ with the vertical
