Could anyone me figure out d and e? how can you find the work of the friction when the force of friction is not given?

Wow !  This is not simple, Shoot, and I give you a lot of credit
and an extra merit badge if you're generally keeping up with it.

I scratched my head for a few minutes, and I think I've got it.
Here's what I think is going on:

KE₁ = KE of the box before pushing
               (1/2) (m) (speed²) = 10 x 2² = 40 joules

KE₂ = KE of the box after pushing 3m
               (1/2) (m) (speed²) = 10 x 4² = 160 joules

The box gained (160 - 40) = 120 J of kinetic energy.

Now look at the cluttered force diagram.
Cat's component of force in the direction of motion is 120N.
That's the part of her force that does the work on the box.
How much work does she do ?

     (force) x (distance) = (120N) x (3m) = 360 joules .

Only 120 J of that energy showed up as increased kinetic energy
of the box.  The other 240J of her hard-earned work was consumed
by friction.

       Work of friction = (Friction force) x (distance)

               240 J        = (friction force)  x  (3 m)

              240 J / 3 m  =  friction 'force'  =  80 N . 

I think that's it.

What I did was:

-- Find the work that Cat did.

-- Find the increase in the kinetic energy of the box.

-- The difference ... the 'missing energy' ... was the work done
by friction in the same distance.

Does this do anything for you ?

Iagree c would be the answer