You calculated the probability of getting exactly $3$ right correctly, and of course the probability of getting all $5$ right. The probability of getting exactly $4$ right is clearly $0$. So we need only deal with the probabilities of $2$ right, $1$ right, and $0$ right.
A small comment about the number of permutations that have exactly $3$ right. Your analysis was good. I would, however, prefer to count as follows. There are $\binom{5}{3}$ ways to pick which three will be right. For each of these ways, we need to swap the other two, so the number is $\binom{5}{3}$. You chose instead the two items that would be swapped. Same count.
For exactly $2$ right, let us count the permutations that have exactly $2$ right. Which $2$ are right can be chosen is $\binom{5}{2}$ ways. For any such choice, that leaves three entries, say $a$, $b$, and $c$, that should be all wrong. Thus $a$ should be in one of positions $b$ or $c$. If $a$ is in position $b$, then $b$ must be in position $c$ (it cannot be in position $a$, for then $c$ would be forced into position $c$, which would make more than $2$ right). It follows that there are $\binom{5}{2}(2)$ permutations in which exactly $2$ are right.
For exactly $1$ right, the one that is right can be chosen in $\binom{5}{1}$ ways. That leaves $4$ items, which should all be in the wrong position. Where does $a$ go? There are $3$ possibilities, $b$, $c$, and $d$. The situation is symmetrical in $b$, $c$, and $d$, so we count the permutations that take $a$ to $b$, and multiply the result by $3$.
So suppose $a$ goes to position $b$. Maybe $b$ goes to $a$, in which case $c$ and $d$ must also get interchanged. This gives $1$ permutation. Or maybe $b$ goes to $c$ or $d$. We count the permutations that take $b$ to $c$, and multiply by $2$. So $a$ goes to $b$, and $b$ goes to $c$. Then $c$ must go to $d$, and $d$ must go to $a$. We conclude that there are $1+2$ allowed permutations such that $a$ goes to $b$. The total is therefore $\binom{5}{1}(3)(3)$.
For exactly $0$ right, we can use the same idea. However, it is easier to subtract the sum of the answers for $5$, $4$, $3$, $2$, and $1$ from $5!$.
Remark: We did crude counting, with essentially no theory. For numbers bigger than $5$, this could become very difficult. There is general theory that gives an answer. For details, look for example at the discussion of derangements in Wikipedia.
For any $m$, let $D_m$ be the number of derangements of a set $S_m$ of $m$ elements, that is, permutations of $S_m$ that leave no element fixed. (Wikipedia uses the notation $!m$ for $D_m$.) There are explicit expressions, and also nice recurrences, for $D_m$. The number of permutations of a set of $n$ elements that leave exactly $k$ elements fixed is then $\binom{n}{k}D_{n-k}$.