(a) 1st bag contains 12 white (W) balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls.
P(two black balls from the 1st bag and two black balls from 2nd bag)
= $(\frac{8}{20}\times \frac{7}{19})\times (\frac{15}{25}\times \frac{14}{24})$
= $\frac{14}{95}\times \frac{7}{20}$
= $\frac{49}{950}$
(b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag)
= $\frac{12}{20}\times \frac{8}{19}\times \frac{15}{25}\times \frac{14}{24}+\frac{8}{20}\times \frac{12}{19}\times \frac{15}{25}\times \frac{14}{24}+\frac{8}{20}\times \frac{7}{19}\times \frac{10}{25}\times \frac{15}{24}+\frac{8}{20}\times \frac{7}{19}\times \frac{15}{25}\times \frac{10}{24}$
= $\frac{42}{475}+\frac{42}{475}+\frac{7}{190}+\frac{7}{190}$
= $\frac{238}{950}$