\begin{equation*}
P(AABBB) = P(A)P(A)P(B)P(B)P(B) = \frac{8}{36} \frac{8}{36} \frac{28}{36} \frac{28}{36} \frac{28}{36}
\end{equation*}
Similarly for the second part.
For the third part, notice that there will be \(\binom{5}{2}\) ways to rearrange 2 A’s and 3 B’s but that each of these will have two 8/36’s and three 28/36’s but just in a different order. Therefore, you will get
\begin{equation*}
10 \cdot \frac{8}{36} \frac{8}{36} \frac{28}{36} \frac{28}{36} \frac{28}{36}
\end{equation*}