Xiao Xin designed the circuit diagram as shown in figure 1 when measuring the current in the circuit with an ammeter. (1) Please use it in Figure 2 according to the circuit diagram drawn by Xiaoxin.

(1) According to the topic, the current starts from the positive pole of the power supply, passes through the switch S 1 and the ammeter, and is divided into two paths. One path passes through L2, then returns to the negative pole through S2, and the other path returns to the negative pole through L 1. As shown in the following figure, according to the data in (2), the ammeter selects a small range:

(2) Only when S 1 is closed, the indicator of ammeter is 0.22A, that is, the current of bulb L2 is 0.22a at this time; When S 1 and S2 are turned off at the same time, the two bulbs are connected in parallel. For L2, the voltage and resistance are constant, so the current of the bulb is constant, that is, I2 = 0.22A At this time, the indication of the ammeter is 0.58A, that is, I=0.58A, so the current of L 1 is: I1= 0.58a-0.22a = 0.36a;

So the answer is: (1) See the above figure; (2) At this time, the currents through L 1 and L2 are 0.36A and 0.22A respectively.