CALCULATIONS :-

i) Produce the linear portion ABC of the graph further to meet Voltage axis at D
Voltage corresponding to point D is Cut - in voltage.
So cut-in voltage is 235 mV.

ii) Let us choose point B on linear portion of graph.
Then R_F = V_F (corresponding to point B) / I_F (corresponding to point B)
= 400 mV / 35 micro A
= 400 x 10^-3 V / 35 x 10^-6 A
= 11.43 k ohms.

iii) Let us apply a signal (a.c voltage) at point B on linear portion of graph with amplitude of 50 mV.
Then r_F = (delta)V_F / (delta)I_F
= (450-350) mV / (47-24) micro A
= 100 x 10^-3 V / 23 x 10^-6 A
= 4.35 k ohms.