All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
(a)
P(faulty)
= P(made by A and faulty) + P(made by B and faulty)
= 0.6(0.05) + 0.4(0.07)
= 0.058

(b)
P(made by A | faulty)
= \frac{\mathrm{P(made~by~A~and~faulty)}}{\mathrm{P(faulty)}}
= \frac{0.6(0.05)}{0.058}
= \frac{15}{29}

(ii)
(a)
P(exactly one of them is faulty)
= 0.058 \times (1 - 0.058) \times 2!
= 0.109272 (exact)

(b)
P(both were made by A | exactly one is faulty)
= \frac{\mathrm{P(both~were~made~by~A~and~exactly~one~is~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}
= \frac{\mathrm{P(one~is~made~by~A~and~faulty,~the~other~is~made~by~A~and~not~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}
= \frac{0.6(0.05) \times 0.6(0.95) \times 2!}{0.109272}
= \frac{1425}{4553}

KS Comments:

Question can be easily solved by drawing a tree diagram. Do take note that we only wrong off NON exact answers to 3sf, so for (iia), we keep the full exact answer.

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