Solution: (1) Note that A, B and C are qualified as events A 1, A2 and A3 after the first firing;

Let e mean that only one piece is qualified after the first firing.

Then: P(E)=P(A 1? A2A3？ )+P(A 1A2？ A3？ )+P(a 1a2a 3)= 45× 14× 13+ 15×34× 13+ 15× 14×23 = 320

The probability that only one product is qualified after the first firing is 320.

(2) Note that after A, B and C are fired twice, the qualified events are A, B and C respectively.

Then: P(A)= 1225, P(B)=920, P(C)=25.

Let f mean that all three products are qualified after twice firing, then: P(F)=P(A? b？ C)= 1225×920×25=54625

∴ The probability that all three products are qualified after two firings is 54625.