a) Since you are told that 4mA is passing through the diode, by using the graph given, you can easily determine that the voltage drop is 0.6V.

b) Since you have a 6V battery and the diode is dropping 0.6V of that, the resistor is dropping the remaining 5.4V.

c) Since the resistor is passing the same 4mA of current that the diode is, Ohm's law tells us that the resistor is R = V/I, or 5.4V / 4mA = 1350 Ohms.

d & e) The figure for these questions is not shown, so I can only assume that it is the same, only the diode is reversed. If the diode is reversed, current will not flow unless the reverse voltage exceeds the reverse breakdown voltage, which the graph indicates is past the -6V mark. This would indicate to me that the diode is dropping the entire 6V (reverse biased), and there is no current flowing.

The characteristic curve is the plot that is shown with the problem. It simply relates the voltage drop across the diode with the current through it. A resistor also has a characteristic curve; it just happens to be linear with a slope of 1/R. We have measured a current of 4mA through the circuit (it may be necessary to convince yourself that the current is the same through all the elements by Kirchoff's current law). If 4mA is going through the diode, we can find how much voltage must be dropped across it by looking at the characteristic curve - find 4mA on the I axis, trace it over to the curve, and find the value off the V axis. In this case, it is about 0.65V.

To find the voltage drop across the resistor, think about Kirchoff's voltage law. We know the sum of the voltages must be 0. In this case, we are considering the ammeter to be ideal - it has 0 volts across it. We know the battery contributes -6V and the diode contributes 0.65V. The only element left is the resistor.

Once we know the voltage drop across the resistor, we have both I and V and can use ohm's law to find its resistance.

If the diode is reversed, no current is flowing (we are assuming it is not broken). If no current is flowing, you can figure out the voltage drop across the resistor through ohm's law. Using the known voltage drops across the battery and the resistor, you can find the drop across the diode using Kirchoff's voltage law.