① The voltmeter is connected in parallel with the bulb, the ammeter is connected in series with the bulb, the sliding rheostat is connected in series with the circuit, and the switch is connected in series with the circuit, as shown in the figure.

(2) according to the position of the ammeter pointer, the current in the circuit may be 0.2A or1a; When the light bulb emits light normally;

The resistance of the sliding rheostat connected to the circuit at this time may be 25 Ω,10 Ω, 5 Ω,

If the indication of ammeter is 1A and the minimum voltage across the sliding rheostat is 1A× 5ω = 5V > 4.5V, then the indication of ammeter cannot be1a.

So the indication of ammeter is 0.2A, that is, the rated current of bulb is 0.2a;

According to U=IR, the voltages across the sliding rheostat may be 5V (greater than the voltage of the power supply, so sliding rheostat is impossible), 2V and1v; ;

According to UL=U-U sliding, the rated voltage at both ends of the bulb may be 2.5V, 3.5V,

According to the specification of the bulb, the rated voltage of the bulb should be 2.5V, and the specification of the sliding rheostat is "20Ω? 2A”，

The rated power of the bulb is P =U = I = 2.5V× 0.2A = 0.5W 。

So the answer is: ① As shown above;

②20Ω? 2A； 2.5; 0.2; 0.5.