current and voltage

It is best to do a simple calculation.

Let R be the resistance of each bulb, and V be the voltage provided by the battery.
Equivalent resistance of the 3 lamps in parallel = R/3
Total resistance of the circuit = R + R/3 = 4R/3
Main current (i.e. current through L1) Io = V/(4R/3) = 3V/4R
Current through each branch in the parallel circuit, Ip = (1/3) x 3V/4R = V/4R

After L3 is burnt,
Equivalent resistance of L2 and L4 in parallel = R/2
Total resistance of circuit = R + R/2 = 3R/2
New main current, Io' = V/(3R/2) = 2V/3R
Current through either L2 or L4, Ip' = (1/2) x 2V/3R = V/3R

Therefore, it is clear that the current through L1 (the main current) has decreased (from 3V/4R tp 2V/3R), but the current through either L2 or L4 has increased from V/4R to V/3R. Hence, L1 will be dimmer and both L2 and L4 will be brighter.

The physical reason is that once L3 is burnt, the equivalent resistance of the parallel circuit (i.e. L2 and L4) will increase. The increase of equivalent resistance shares a higher proportion of voltage from the battery than before. This increase in voltage causes both L2 and L4 to become brighter.

Since L2 and L4 now shares a higher proportion of battery voltage, the voltage across L1 will decrease, leading to a dimmer result.

For quicker thinking,
you may think that,
V supplied is constant
For L3 is burnt, Resistance of combining of L2,L3,L4 increases,
According to V proportional to R, Combining resistance receive more V,
For V_total = V_1 + V_combining => V_1 decrease
then L1 is dimmer

2012-03-22 00:02:19 補充：
Since V supply to combing resistor increases,
then for less sharing of p.d., then L2, L4 are brighter since more energy are supply to them