Part c if exena applies a force of 220 n at the edge of the door and perpendicular to it, how much time t does it take her to close the door?

The initial formula is J·α = F·w but we need to find the angular acceleration, so the other formula is α = F·w/J 

where 
J moment of inertia, α angular acceleration, w width of the door (= length of lever arm of the force F) 

Moment of inertia of plane rectangular object, rotating around axis through on edge (see link) 
J = (1/3)·m·w 
since mass equals weight W divided by acceleration due to earth gravity g 
J = (1/3)·(W/g)·w² 

So the angular acceleration of the door is: 
α = F·3·g / (W·w) 
= 220N · 3 · 9.81m/s² / (750N · 1.25m) 
= 6.90624 s^-2 

Integrating twice with initial conditions 
ω(t=0) = 0; and

α(t=0) = 0 

look for the angular displacement of the door: 
α = dω / dt will be ω = ∫ α dt = α·t 
ω = dφ / dt will be φ = ∫ ω dt = ∫ α·t dt = (1/2)·α·t² 

So time elapsed until door closed at an angle of 
φ' = 90° = (1/2)·π 
is 
t' = √( 2·φ'/α ) 
= √( π / 6.90624 s^-2 ) 
= 0.67s