3.3) I'm not sure whether it is correct or not. I'm am confused because of the Applied Force being at an angle. Am I only working out Fx and would...

Hello!

There are some mistakes in your solution. First, the transition from `F_f` to `mu*mg` is incorrect. We know that `F_f=mu*N,` where `N` is the reaction force. But in this case `N!=mg` because `F` acts partly upwards. We have to consider the both projections, vertical and horizontal.

For the vertical axis we obtain  `N-mg+F*sin(alpha)=0`

and from the horizontal  `F*cos(alpha)-F_(f) =F*cos(alpha)-mu*N=0.`

Express `N` from the first equation and substitute it into the second:

`N=mg-Fsin(alpha),`

`Fcos(alpha)=mu(mg-Fsin(alpha))=mu mg-F mu sin(alpha).`

So  `F(cos(alpha)+mu sin(alpha))=mu mg`  and the final formula is

`F=(mu mg)/(cos(alpha)+mu sin(alpha)).`

Now recall that `alpha`=20°, `mu=0.4,` `m`=50 kg and `g`=9.8 `m/s^2,` and compute:

`F approx (0.4*50*9.8)/(0.94+0.4*0.34) approx 196/1.08 approx` 181 (N). This is the answer for 3.3. Now you can easily solve 3.4 and 3.5.

Note also that 0.4*50*9.8 = 196 and not 16 as you wrote (maybe you simply omitted 9).