I subscribe to the a = (6.1ihat - 3.95 jhat) m/s, and also to force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N
but from there you start losing me:
If you go one step to your right (##\hat\imath##) and one step forward (##\hat\jmath##), the result is that your displacement is √2 steps at 45 degrees.
It's called vector addition. The two forces have to be added as vectors -- if you want to add them at all. But in this exercise you don't want that. They give velocities as ##\ ( a\; \hat\imath + b\; \hat\jmath\ )\ ##, so an answer in the same terms should be OK.
The gravity force that works in the ##\hat\jmath## direction is ##\ m\;\vec g\ = 3.7 * -9.81\; \hat\jmath\ ##. If the net force is ##\ -14.6\; \hat\jmath\ ## N (I hope you didn't forget the minus sign ?) then the applied force in the y-direction must be quite positive.
This whole escapade can take place on an inclined plane or in thin air -- if the exercise tekst says "
Two forces act on the mass: gravity and an unknown applied force" then that is enough to calculate that one unknown force. Which is all the exercise wants you to determine.