U0=8V, the small bulb is "6V, 3W", and the voltmeter shows 4V after S is closed.

I didn’t see the picture, but if the resistance of the light bulb is calculated as a fixed value (not considering the different resistance values ??when the brightness is different), the power P is U square divided by R, and the resistance R is 36/3 =12Ω. The actual power is still based on U square/R as 16/12 =1.3333W.

To make the small lamp glow normally, you need to reduce the effective resistance of the sliding rheostat (no picture, adjust it yourself according to the picture) because It is connected in series, so the current on the small lamp should be P=UxI. The current I is 0.5A, so the current in the resistor is 0.5A. The voltage across the resistor is 8-2 =6V U=IxR so the resistance of the varistor is 4Ω.