To set up Bayes' rule, first think the following way : FIX a coin. Now look at the probability of success.
Suppose a coin has probability $p$ of coming heads. From independence of successive flips, we conclude that the probability of the first three flips looking like $HTT$ is $p(1-p)^2$.
Now, you have three such coins, with $p=0.6,0.5,0.3$. So we can calculate the probability of , each one being FIXED, the first three flips looking like HTT, and these are (easy to see) respectively $0.096,0.125,0.147$.
Now, one of these coins are chosen uniformly at random. So, let us call the coins as coin $1,2,3$ in the order $p=0.6,0.5,0.3$.
Now, Bayes' rule would basically say this : the probability that you get HTT is the sum of the following:
1) The probability that you choose coin $1$, and it comes HTT.
2) The probability that you choose coin $2$, and it comes HTT.
3) The probability that you choose coin $3$, and it comes HTT.
What are these? Well, the choice is independent of the flipping, and is uniform, hence we have the probability of each one being chosen is $\frac 13$. Finally, the complete answer is $\frac{0.096+0.125+0.147}{3} = \frac{46}{375}$.
Now, the second question is asking this : given that HTT has occurred, what is the probability that we chose coin $2$. But then, coin $2$ is just case $2$ above, which is just the probability that with the second coin fixed, we get HTT.
So the answer is just the probability of getting HTT with coin $2$, divided by the probability of getting HTT with any of them(but here, there is no choice to be made, so we do not need to multiply by $\frac 13$). The numerator is then just $0.125 = \frac 18$ while the denominator is $\frac {46 \times 3}{375}$. Now, you can simplify and get $\frac{375}{1104}$.