Total number of possible rolls: $6^8 = 1679616$.
I see two possible interpretations of the question. In the following, I answer both of them.
Interpretation 1:
There are exactly 3 twos or exactly 3 threes
Number of rolls with exactly 3 twos: $\binom{8}{3}\cdot 5^5 = 175000$.
(Explanation: $\binom{8}{3}$ is the number of choices of the positions of the twos. The factor $5^5$ comes from the remaining $5$ dice, which have $5$ possibilities (every number except two) each.)
Number of rolls with exactly 3 threes: $\binom{8}{3}\cdot 5^5 = 175000$.
Number of rolls with exactly 3 twos AND exactly 3 threes: $\binom{8}{3}\binom{5}{3}\cdot 4^2 = 8960$.
So by the sieve formula, the number of rolls with exactly 3 twos OR exactly 3 threes is $175000 + 175000 - 8960 = 341040$.
Thus, the probability is
$$
\frac{341040}{1679616} \approx 20.3\%.
$$
Interpretation 2: Exactly 3 dice show the digit 2 or 3
The number of such rolls is $\binom{8}{3}\cdot 2^3 \cdot 4^5 = 458752$.
So the probability is
$$\frac{458752}{1679616} \approx 27.3\%.$$
This question is equivalent to, what is the probability any particular sequence will appear if a dice is rolled $6$ times, the fact that this particular sequence happens to be $1,2,3,4,5,6$ is irrelevant. Hence there is a $\frac{1}{6}$ chance a $1$ will be rolled first, $\frac{1}{6}$ chance a $2$ will be rolled second, $\frac{1}{6}$ that a $3$ will be rolled third, etc. Therefore the probability of the sequence appearing is
$$\frac{1}{6}\cdot\frac{1}{6}\cdot \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} = \frac{1}{46656}.$$
Best Answer
Let's look at one possible way:
$$x_1, 5, x_2, 5, x_3$$
The $x_i$ stand for "anything else but 5." The probability of this event is $(\frac{1}{6})^2(\frac{5}{6})^3$. So, it can happen in this particular way OR (this is the key word) another way corresponding to all the other ways to arrange the fives. How many ways are there to arrange two objects in a set of five?
The answer is $\binom {5}{3}$. Hence, the probability of your event is $$\binom {5}{3}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3.$$