Answer: Option D. -> > 500 N for time △ t1 and △ t3, and 500 N for △ t2,
:
D
So when scooter is accelerating the man is also accelerating along with it so friction from the scooter's seat on man is making him move forward.
Also there is a Normal force

N = mg = 500 N
So Net force while accelerating will be $\sqrt{{\left(500\right)}^2+{\left(f_r\right)}^2}$> 500
Same is true about decelerating

So while accelerating or decelerating the force on man by scooter is more than 500N. When the scooter is going with constant velocity, the man is also going with constant velocity. So 0 acceleration in horizontal direction so no net force in horizontal direction so no friction.
Only Normal N = mg = 500

Answer: Option B. -> 10N
:
B
u = 10m/s
v = 0
s = 50
$v^2$- $u^2$= 2as
a = -1 m/ $s^2$
Friction from is providing this deceleration of the block

fr = ma
fr = 10 $\times$1
fr = 10N
so friction is 10N in backward direction.

Answer: Option B. -> decrease 
:
B

Now if the angle made by $F_c$and the vertical decreases then horizontal componenet decreases

As you can see the horizontal component decreases, Which in this case turns out to be frictional force.
Yes the magnitude of frictional force depends upon the angle made by Contact force and the Vertical/Horizontal