a) The computation takes a while. It may be useful to draw a tree in order not to lose track of the possibilities. Initially, we toss $3$ dice. We get $0$, $1$, $2$, or $3$ $6$'s. Then, depending on the outcome, we toss a certain number of coins. So from the "start" position, there are $4$ branches, corresponding to the number of $6$'s obtained.
If $3$ dice are rolled, then the probability of $k$ $6$'s is equal to $\binom{3}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{3-k}.$ In particular, the probability of $0$ $6$'s is $\frac{125}{216}$, the probability of exactly $1$ $6$ is $\frac{75}{216}$, the probability of exactly $2$ $6$'s is $\frac{15}{216}$, and the probability of $3$ $6$'s is $\frac{1}{216}$.
If we have $0$ $6$'s, we toss $0$ coins, so for sure (probability $1$) we get fewer than $2$ heads. If we get $1$ $6$, then we toss $1$ coin, and again for sure we get fewer than $2$ heads. If we get $2$ $6$'s, we toss $2$ coins, and the probability of fewer than $2$ heads is easily computed to be $3/4$. If we get $3$ $6$'s, we toss $3$ coins. By symmetry the probability of fewer than $2$ heads is then $1/2$. It follows that the overall probability of fewer than $2$ heads is $\frac{125}{216}\cdot 1 + \frac{75}{216}\cdot 1 +\frac{15}{216}\cdot \frac{3}{4}+\frac{1}{216}\cdot\frac{1}{2}.$ This may simplify to $\dfrac{847}{864}$. But check the arithmetic!
b) Let $E$ be the event that exactly $2$ sixes were obtained, and let $L$ be the event we got fewer than $2$ heads. We want $P(E|L)$.
There are various formulas that we could now use. We go for the simplest one, though using it is a bit less mechanical than the full Bayes formula. Note that $P(E|L)P(L)=P(E \cap L).$ This formula comes straight from the the definition of conditional probability. From the above formula, we see that we will be essentially finished once we find $P(E\cap L)$ and $P(L)$.
The probability $P(L)$ that we get fewer than $2$ heads has been computed in part (a). So we only need to find $P(E \cap L)$. In the solution of part (a), we already saw that $P(E)=\binom{3}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^1=\frac{15}{216}=\frac{5}{72}.$ Given that we got exactly $2$ $6$'s, we got to flip the coin twice, and thus the probability of fewer than $2$ heads is $3/4$, as shown by @Yuval Filmus. It follows that $P(E\cap L)=P(L|E)P(E)=\left(\frac{3}{4}\right)\left(\frac{5}{72}\right)=\frac{5}{96}.$ Now put the pieces together. I think that $P(E|L)=\dfrac{45}{847}$. Do check the arithmetic!