When you roll 2 dice there are 36 equally likely options, starting at double one, $(1,1)$, and going all the way to double six, $(6,6)$. Of these, there are 11 equally likely ways to roll a $5$, $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3),(5,4),(5,6)$, so the odds of rolling a 5 is $\frac{11}{36}$.
Similarly there are 18 equally likely ways to roll an odd number (I'll let you think about which combinations these are), so the odds of rolling an odd number are $\frac{18}{36}=\frac{1}{2}$.
At this point you may think the odds of rolling a $5$ or and odd number should be $\frac{11}{36}+\frac{18}{36}=\frac{29}{36}$, however this is not the case. Note that six of the rolls contain both a $5$ and are odd:
$(2,5), (4,5), (6,5), (5,2), (5,4),(5,6)$
If we just add the two probabilities, we'll count these situations twice, even though they're no more likely to occur than any other combination! In reality there are only $11+18-6=23$ dice rolls which contain a $5$ or are odd (see if you can list them), and so the odds of rolling either a $5$ or an odd combination is $\frac{11}{36}+\frac{18}{36}-\frac{6}{36}=\frac{23}{36}$.
This is a simple application of the "principle of inclusion and exclusion", which you may want to look up.
Most of your work is correct.
There are two exceptions.
If the three dice are sequential, then they must assume one of the following sets of values $\{1, 2, 3\}$, $\{2, 3, 4\}$, $\{3, 4, 5\}$, or $\{4, 5, 6\}$. For each of the four sets, there are $3! = 6$ permutations which result in the same set of values. Thus, the probability that the dice are sequential is
$$\frac{4 \cdot 3!}{6^3} = \frac{24}{216} = \frac{1}{9}$$
If events $A$ and $B$ are independent, then $P(A \cap B) = P(A)P(B)$. If we let $A$ be the event that the sum of the dice is $8$ and $B$ be the event that none of the dice rolled show a 2, then
\begin{align*}
P(A) & = \frac{21}{216} = \frac{7}{72}\\
P(B) & = \frac{125}{216}\\
P(A \cap B) & = \frac{9}{216} = \frac{3}{72}
\end{align*}
As you can check, $P(A \cap B) \neq P(A)P(B)$. The events are dependent because if none of the dice rolled show a 2, you are less likely to get a sum of $8$ since it becomes more likely that the sum of the numbers shown on the dice will be too large.
Best Answer
You will get a prime product if exactly one of the faces is prime and the rest are all 1. The probability (binomial distribution) that you will have a prime product with n dice is $\displaystyle \binom{n}{1}.(\frac{1}{2})^1.(\frac{1}{6})^{n-1}$
So you need to find $\displaystyle \frac{1}{2}\sum_{n=0}^\infty n(\frac{1}{6})^{n-1}$
Replace $\frac{1}{6}$ by $x$. Now consider the geometric sum $1 + x + x^2 + x^3 + ... x^n + ...$. What happens when you differentiate it term by term wrt $x$? Especially consider what happens to the general term $x^n$.
Now find an expression for the geometric sum in terms of $x$ and differentiate that wrt $x$. Finally substitute $x=\frac{1}{6}$. That should give you the required sum.
Post a comment if you still have difficulty.