Answer

First note that it is relatively easy to compute probability that max is smaller than some number, since that simply means all rolls are smaller than that number, i.e. $$P(max\leq 5)=P(X_1\leq 5)\cdot P(X_2\leq 5)\cdot ... \cdot P(X_8\leq 5)=(5/6)^8$$This is because the dice are identical and for one to be smaller or equal to 5 there are 5 possibilities (numbers 1 to 5). Okay, so it is straightforward to compute $P(max \leq k)$ for any k. So how can we use this to find our answer?

Well we need a way to express maximum being exacly 5 in terms of statements of the form 'maximum is smaller or equal to k'. Well let's simply set the upper and lower bounds: maximum has to be smaller or equal to 5, and it has to be greater than 4. Greater than 4 is the opposite of smaller or equal to 4. In turn, we can write this as: $$\{max=5\}=\{max\leq 5\}\land\{max\gt 4\}=\{max \leq 5\}\backslash \{max \leq 4\}$$Note that event '$max \leq 5$' is implied by '$max \leq 4$' since if you're below 4 you're definitely below 5. Therefore the event '$max \leq 4$' is fully contained in '$max\leq 5$'. That means we can compute the chance of max below 5 but not max below 4 as: $$P(max\leq 5)-P(max \leq 4)$$As we've already seen, these terms can be computed easily. Denoting X a single dice toss, we have: $$P(max\leq 5)=P(X \leq 5)^8=\left(\frac{5}{6}\right)^8$$ and similarly: $$P(max\leq 4)=P(X \leq 4)^8=\left(\frac{4}{6}\right)^8$$Therefore we can find the solution as: $$\left(\frac{5}{6}\right)^8-\left(\frac{4}{6}\right)^8\approx 0.19$$