Answer
65km/h, 58$^o$ west of north.
65km/h, 32$^o$ south of east.
Work Step by Step
Call east the positive x direction and north the positive y direction.
Let 1 denote Car 1, 2 denote Car 2, and s the street.
The pair “12”, for example, represents Car 1’s motion relative to Car 2, i.e., as seen by Car 2.
$$ \vec{v_{1s}} = \vec{v_{12}} + \vec{v_{2s}} $$
$$(0, 35 km/h) = \vec{v_{12}} + (55 km/h, 0)$$
Evidently, $\vec{v_{12}} = (-55 km/h, 35 km/h)$. As expected, this is north and west (58$^o$).
$\vec{v_{12}} = \sqrt {(-55 km/h)^2 + (35 km/h)^2} = 65.2km/h \approx65km/h.$
The velocity of car 2 relative to car 1 is exactly the opposite of this.
$\vec{v_{21}} = (55 km/h, -35 km/h)$
$\vec{v_{21}} = \sqrt {(55 km/h)^2 + (-35 km/h)^2} = 65.2km/h \approx65km/h$ $and$ $32^o south of east$
