HINT
The way you approach each sub-problem is actually completely correct. These are binomial distributions and e.g. your formula
$$3 \times \frac16 \times (\frac56)^2$$
can be found in the wikipedia article in the form
$${n \choose k} \times p^k \times (1- p)^{n-k}$$
for the case of $n=3$ dice rolls, you want exactly $k=1$ instance of outcome "2", and the probability of that outcome in any specific trial being $p=1/6$. (BTW the same formula, if you plug in $p=1/2$ would give you $3/8$ for the "exactly $1$ Head in $3$ coin tosses" example.)
However, as you said, ${3 \over 8} \times {75 \over 216}$ is the prob of getting exactly one Head and exactly one "2". You want one or the other, not both. This needs to be counted as:
$$Prob(\text{exactly one Head}) \times Prob(\text{no 2}) + Prob(\text{no Head}) \times Prob(\text{exactly one 2})$$
Each of the $4$ individual probabilities can be figured out easily.
