I = I 1 = u 1r 1 = 12v 24ω= 0.5A；

(2) When S3 is only closed: I2 = Ur2 =12v100ω = 0.12a.

P2 = UI2 = 12v×0. 12A = 1.44 w；

(3)① Only the key S2 is closed, and the resistor R 1 is connected in series with R3.

When u1= 3v: u3 = u-u1=12v-3v = 9v,

I = u 1r 1 = 3v 24ω= 0. 125 a，

R3 = U3I = 9v 0. 125 a = 72ω；

② Only turn off the button S2, and the resistor R 1 and R3 are connected in parallel.

I 1 = ur 1 = 12v 24ω= 0.5A，

When I = 3a: i3 = I-I1= 3a-0.5a = 2.5a,

R3 = UI3 = 12v 2.5a = 4.8ω；

When I=0.6 ampere: i3 = I-I1= 0.6a-0.5a = 0.1a,

R3 = UI3 = 12v 0. 1A = 120ω；

Answer: (1) When only the S 1 key is closed, the voltage indicator is 12 volts, and the ammeter indicator is 0.5A:

(2) When only key S3 is turned off, what is the resistance? The electric power consumed by R2 is1.44 w;

(3) The possible resistance values of constant resistance R3 meeting the above requirements are 72 Ω, 4.8 Ω,120 Ω.