Well, if you discount the rolling resistance of the tires/drive train, at the fixed output power of the engine when traveling 60 MPH, the force vectors are all in the same horizontal plane: The force of air resistance is matched by the engine's output. For a given object, air resistance is proportional to velocity squared.
On the inclined plane, you must break the velocity vector into the horizontal and vertical components. (At 45 degrees , the horizontal and vertical speed components are equal, and both equal the velocity divided by the square root of 2.)
On an inclined plane, the fixed engine output must be sufficient to lift the car's mass against the force of gravity. Since overcoming the drag at 60 MPH would be on the order of 20 HP/15KW, it is entirely possible that this 'fixed' output would not be enough to lift the car at all, unless the speed dropped to near zero. (in real life, given most vehicles gear reduction range, it's probable that a car would not be able to go up this steep an incline at all.) At 45 degrees, the force vector due to gravity is split into the horizontal and vertical components by dividing mass by sqrt(2), also. And, the air resistance vector, which is parallel to the plane, would need to be included needs to be included in the solution as well.
I really think you should take a look at the force diagrams and explanations at the lesson on inclined planes
here. Diagrams are helpful in understanding the forces involved.