If two randomly chosen envelopes are wrongly addressed, then the other two envelopes can only be correctly addressed if the two persons who receive the randomly chosen envelopes can get the letter that is meant for them by switching. Person $A$ must have received the letter for person $B$ and vice versa. If not then person $C$ or $D$ will receive a letter meant for $A$ or $B$ and more than $2$ letters are wrongly addressed.
Let's say that order $\left(1,2,3,4\right)$ stands for a fully correct
addressing. The first two letters are wrongly addressed if
we are dealing with for instance $\left(2,4,.,.\right)$. Counting
tells you that there are $14$ of such cases. The first two letters
are 'switched' in the cases $\left(2,1,3,4\right)$ and $\left(2,1,4,3\right)$
and in only case $\left(2,1,3,4\right)$ there are exactly $2$ wrongly
addressed letters. If the first two letters are not switched then
there are more than $2$ wrongly addressed letters as argued above. We find
a probability of $\frac{1}{14}$ that there are exactly $2$ letters
are wrongly addressed under condition that the first two are wrongly
addressed.
Let us count the number of ways that letter A can be in its correct envelope, which we also call A, while none of the others is in a correct envelope.
So B must go to C or D ($2$ choices).
If B goes to C, then C cannot go to B, else D would have to go to D. So C must go to D, and the fate of D is determined. So there is only $1$ way that B can go to C.
The same is true if B goes to D. So overall there are $2$ choices where A is the "fixed point."
That gives a total of $8$ possibilities. Divide by $4!$.
Remark: We could also solve this in a much more general way. We have $n$ letters, and want to find the probability that exactly $k$ of them end up in the correct envelope. The $n$ letters can be permuted in $n!$ ways. As in the answer above, that will be the denominator.
For the numerator, the lucky letters that end up in the right envelope can be chosen in $\binom{n}{k}$ ways. The unlucky $n-k$ letters can be all put into wrong envelopes in $D(n,k)$ ways, where $D(n,k)$ is the derangement number. For more about counting Derangements, please see Wikipedia. It uses the notation $!w$ for the number of derangements of $w$ objects.
Best Answer
Note the use of the word exactly. You did not take that into account.
Probability that exactly three letters are placed in the correct envelopes
There are indeed $\binom{5}{3}$ ways to select which three letters are placed in the correct envelopes. That leaves two envelopes in which to place the remaining two letters. Those letters must each be placed in the other envelope, which can only be done in one way. Hence, the probability that exactly three letters go in the correct envelopes is $$\frac{1}{5!} \cdot \binom{5}{3}$$
Probability that exactly two letters are placed in the correct envelopes
There are $\binom{5}{2}$ ways to select which two letters are placed in the correct envelopes. None of the remaining letters go in the correct envelopes. There are just two ways to do this. \begin{align*} &\color{red}{1}, \color{red}{2}, \color{red}{3}\\ &\color{red}{1}, 3, 2\\ &2, 1, \color{red}{3}\\ &2, 3, 1\\ &3, 1, 2\\ &3, \color{red}{2}, 1 \end{align*} Hence, there are $\binom{5}{2} \cdot 2$ favorable cases.
We can use the Inclusion-Exclusion Principle to see why.
There are $3!$ ways to permute the three letters. From these, we must subtract those cases in which at least one letter is placed in the correct envelope.
There are three ways to select a letter to be placed in the correct envelope and $2!$ ways to place the remaining letters.
There are $\binom{3}{2}$ ways to select two letters to be placed in the correct envelopes and $1!$ ways to place the remaining letter.
There are $\binom{3}{3}$ ways to select three letters to be placed in the correct envelopes and $0!$ ways to place the (nonexistent) remaining letters.
By the Inclusion-Exclusion Principle, the number of ways to place the remaining three letters so that none of them is placed in the correct envelope is $$3! - \binom{3}{1}2! + \binom{3}{2}1! - \binom{3}{3} = 6 - 6 + 3 - 1 = 2$$ as claimed.
Therefore, the probability that exactly two letters are placed in the correct envelopes is $$\frac{1}{5!} \cdot 2\binom{5}{2}$$
Probability that none of the letters is placed in the correct envelopes
This is a derangement problem. An Inclusion-Exclusion arguments shows that the number of ways of placing none of the five letters in its correct envelope is
Dividing that number by $5!$ gives the desired probability.