Physics:why reading of A1 is 0.6, i don't understand how to do, thanks?

First, note the voltage across A₂ is zero. That's because, by symmetry the voltage  at the top of A₂ is the same as the voltage at the bottom.  So no current flows through A₂ and the circuit behaves the same as if A₂ were not there.

The top pair of resistors are in series giving a total of 20Ω.
The bottom pair of resistors are in series giving a total of 20Ω.

The two pairs are in parallel so their total R is:
1/R = 1/20 + 1/20 = 1/10
R = 10Ω
(When you get familiar with circuits you will know the answer is 10Ω without any calculations needed.)

Current I = V/R = 6/10 = 0.6A

So ammeter A₁  reads 0.6A (don't forget unit)

Ammeter has negligible resistance, and thus the resistance of the ammeter A₂ is negligible.

Equivalent of the four resistors, R
= 1 / {[1/(10 + 10)] + [1/(10 + 10)]} Ω
= 10 Ω

Ohm's law: V = IR
Reading of ammeter A₁
= V/R
= 6/10 A
= 0.6 A

R = 20*20/(20+20) = 10
i = V/R
i = 6/10
I=0.6 A