(1)(2) According to the meaning of the question, the current when the two bulbs emit light normally is I1= Pu1=1w1.5v ≈ 0.7a; I2=PU2= 1W3V≈0.3A, and the maximum current does not exceed1a; ;

As can be seen from the figure, the measuring range of ammeter should be 0 ~ 0.6a, indicating 0.24A；;

According to the sliding vane of the sliding rheostat near the midpoint, we can know that the resistance of this bulb is close to 12ω× 12 = 6ω or 20ω× 12 = 10ω.

If the rated voltage of Xiao Wei bulb is 1.5V, the resistance of sliding rheostat R=UI=4.5V? 1.5v 0.24 a≈ 12.5ω；

If the rated voltage of Xiao Wei bulb is 3V, the resistance R=UI=4.5V? 3v 0.24 a≈6.25ω；

According to the above analysis, Xiao Wei uses 12ωA sliding rheostat, and the rated voltage of the bulb is 3V, so the rated power of Xiao Wei bulb is p1= u1I1= 3v× 0.24a = 0.72w;

The rated voltage of Xiao Li's bulb is 1.5V, and the voltmeter is connected in parallel with the sliding rheostat, so the wiring is wrong;

(3) According to the meaning of the question and the above analysis, the rated voltage of the bulb used by Xiaoli is 1.5V, and the reading of its voltmeter is 3V, so the current in the circuit at this time I = 0.96W3V = 0.32A

Then p = ui =1.5v× 0.32a = 0.48w. 。

So the answer is:

(1) Xiao Li;

(2) A; 0.72;

(3) From the meaning of the question and the above analysis, it can be seen that the rated voltage of the bulb used by Xiaoli is 1.5V, and the indication of its voltmeter is 3V, so the current in the circuit at this time I=P rheostat U rheostat = 0.96W3V = 0.32A

Then p2 = u2i2 =1.5v× 0.32a = 0.48w. 。