Without Replacement: You shuffle the deck thoroughly, take out three cards. For this particular problem, the question is "What is the probability these cards are all Kings."
With Replacement: Shuffle the deck, pick out one card, record what you got. Then put it back in the deck, shuffle, pick out one card, record what you got. Then put it back in the deck, pick out one card, record what you got. One might then ask for the probability that all three recorded cards were Kings. In the with replacement situation, it is possible, for example, to get the $\spadesuit$ King, or the $\diamondsuit$ Jack more than once.
For solving the "without replacement" problem, here are a couple of ways. There are $\binom{52}{3}$ equally likely ways to choose $3$ cards. There are $\binom{4}{3}$ ways to choose $3$ Kings. So our probability is $\binom{4}{3}/\binom{52}{3}$.
Or else imagine taking out the cards one at a time. The probability the first card taken out was a King is $\frac{4}{52}$. Given that the first card taken out was a King, the probability the second one was is $\frac{3}{51}$, since there are $51$ cards left of which $3$ are Kings. So the probability the first two cards were Kings is $\frac{4}{52}\cdot\frac{3}{51}$. **Given that the first two were Kings, the probability the third is is $\frac{2}{50}$. So the desired probability is $\frac{4}{52}\cdot\frac{3}{51}\cdot \frac{2}{50}$.
Remark: We could solve the same three Kings problem under the "with replacement" condition. (You were not asked to do that,) The second approach we took above yields the answer $\left(\frac{4}{52}\right)^3$. Since we are replacing the card each time and shuffling, the probability of what the "next" card is is not changed by the knowledge that the first card was a King.
You are assuming that order doesn't matter in the second case, but this is the wrong assumption. Order definitely matters. You have, in the case of $n=6$ a sample space of $6^2$ Even though when you finally pull both tags out, a $(2,1)$ is the same as $(1,2)$, these are still different events and must be treated differently. Since this is the case, let's look at $n=6$. All consecutive numbers then would be;
$$(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)$$
So here you have 10 events that are possible, not 5. And your sample space is 36... Therefore, your probability is
$$\frac{10}{36}=\frac{2\cdot5}{6^2}=\frac{2(6-1)}{6^2}$$
And this makes sense. You can only have $n-1$ consecutive pairs, since the $n$-th pair would be $(n,1)$ which are not consecutive, and our sample space consists of $n^2$ events. Since there are two ways to get consecutive integers, the formula is
$$P(\text{consecutive numbers with replacement of n tags})=\frac{2(n-1)}{n^2}$$
Also, in the first case, again, you are making the faulty assumption that $(1,2)$ is the same as $(2,1)$ and I think that you make this assumption because the probability is correct under both assumptions, i.e., yours, and the correct assumption. Why?
The sample space under sampling without replacement is in the case of $n=6$ is $6\cdot5=30$. This is the case because pairs such as $(1,1), (2,2)$ are impossible without replacement. However, you can still get the ten pairs of consecutive numbers listed above, so therefore, under the correct assumption
$$P(\text{consecutive numbers without replacement of n tags})=\frac{2(n-1)}{n(n-1)}=\frac2{n}$$
If $n=6$, you get your probability is $\frac1{3}$, which is what you got under your faulty assumption. Hope this helps shed a little light.
Best Answer
I will try to reach unto you a new way of thinking.
Suppose that all parts are given randomly a distinct number from $\{1,2,\cdots,25\}$.
What is the probability that e.g. part numbered $17$ is one with excessive shrinkage?
In (a) and (b) you are somehow giving numbers too. However in (a) you stop this if after giving $2$ numbers (the first selected gets number $1$, the second gets number $2$...) and in (b) after giving $3$ numbers. Nothing stops you from going on with the numbering until you reach $25$.
You can also think of $25$ numbered parts. Then $5$ of the numbers are randomly chosen and get the stamp "with excessive shrinkage". What is the probability then number $17$ is among those $5$? Same answer.
Take care of it that you master this way of thinking. It will bring you great profit in the study of probabilities.