Hint: You will get no more than 10 sixes if you get no sixes or one six or two sixes or three sixes or ... or nine sixes. Since these possibilities are mutually exclusive, you can add the individual probabilities to get the total probability.
Once you've solved (b), think how you can use that answer to solve (c).
Record the outcome of the rolling as a string of length $8$, made of symbols chosen from $\{1,2,3,4,5,6\}$. There are $6^8$ such strings, all equally likely if the die is fair and rolled fairly.
Now we count the number of good strings, where good means fitting the specifications in the title.
The location of the digit which is different from $3$, $1$, or $6$ can be chosen in $8$ ways. For each choice of location, there are $3$ ways to choose which one of the three possibilities the "odd" digit will be. Thus there are $(8)(3)$ ways to decide on the location and type of the "odd" digit.
For every way of doing the above two tasks, there are $7$ empty slots. There are $\frac{7!}{2!3!2!}$ ways of filling them with two $3$'s, three $1$'s, and two $6$'s. That gives a total of $(8)(3)\left(\frac{7!}{2!3!2!}\right)$ good strings. Divide by $6^8$ for the probability.
We can count the number of ways of filling the $7$ slots more slowly. The location of the $3$'s can be chosen in $\binom{7}{2}$ ways. For each of thse choices, the location of the $1$'s can be chosen in $\binom{5}{3}$ ways, for a total of $\binom{7}{2}\binom{5}{3}$. That gives the expression $(8)(3)\binom{7}{2}\binom{5}{3}$ for the number of good strings.
Remark: There are other ways to organize the count. For example, let us make an $8$-letter string, made up of two $3$'s, three $1$'s, two $6$'s, and an $x$, to act as a placeholder for the "odd" number. There are $\frac{8!}{2!3!2!}$ ways to do this. Multiply by $3$, because $x$ can be replaced by any of $3$ values.
Best Answer
alternative approach
$E$ is the event of a 1 or a 6
$p(E) = 1/3$.
Let $p = 1/3.$
Let $q = 1 - p$.
General formula for exactly $k$ successes in $n$ tries, where
$n \in \mathbb{Z^+}$ and $k \in \{0,1,\cdots, n\}$ is
$$\binom{n}{k}p^kq^{(n-k)}$$
Here, $n=10$ and $k=4.$