Interesting result. How did you measure voltage and current?
The yellow graphs follows the power supply - probably similar to a sine wave (minus some offset for the diode).
No, as the additional current is needed to charge the capacitor. It discharges itself via the lamp when the power supply is at a negative voltage.
One guess: The capacitor increases current and therefore the average load of the lamp - it gets hotter, and its resistance might increase. If the power supply is not perfect and its voltage depends on the current, this might lead to a higher peak voltage. I would not expect this, but something has to explain your observations and I don't see anything better at the moment.