Let me explain again, for the benefit of Young and komalbarun...
If you have a voltage divider let's say the 7 ohms and 5 ohms suggested by komakbarun.
If you put 12 volts across it, you will get 5 volts across the 5 ohm resistor.
Now let's put a load on it. Let's assume our load is 5 ohms (that would be similar to a 5V 5W lamp).
The voltage the load sees is now 12* ((5||5)/((5||5) + 7)
5||5 is 2.5 ohms so
V = 12*(2.5/(2.5+7)) = 12* (2.5/9.5) = 12* (0.2632) = 3.158 volts.
Clearly it's not 5V.
My solution would be to put a 7 ohm resistor in series with the bulb, and the voltage across the bulb would be...
5V
But apparently it's far too complex to get this right...
Now, my solution depends on Young answering the question of the lamp rating in Watts, or the current in Amps, which he has not, despite being asked more than once.
Now, it's also possible to allow for the resistor in parallel with the bulb.
You could have a resistor of 3.5 ohms and a resistor of 5 ohms.
Unloaded, this would produce a voltage of 7.06V, but with the bulb in place you would again get 5V.
However this is simply wasting the power flowing through the 5 ohm resistor.