Hints:
Exactly four right is obviously impossible. If four get the right hat, the fifth must also. The probability is $0$.
All wrong is a famous problem, the problem of counting Derangements.
The Wikipedia article is quite helpful. Note that there are $5!$ equally likely ways that the hats can be distributed among the $5$ people. From the formula in the article, you will find there are $44$ derangements of $5$ objects, so the required probability is $\dfrac{44}{5!}$.
For the probability that more than one gets the right hat, it is easier to find the probability that one or fewer gets the right hat. The probability of none is the answer to (b). For the probability of one, the lucky person can be chosen in $5$ ways. Multiply this by the probability none of the remaining $4$ get the right hat. This again is a derangements problem. For $4$, you can easily list and count the derangements. There are $9$. So the probability exactly one person gets the right hat is $\dfrac{45}{5!}$.
Thus the probability of (c) is $1-\dfrac{44}{5!}-\dfrac{45}{5!}$.
Here are some observations. Start with the species $\mathcal{Q}$ of
permutations with fixed points marked, so that we have
$$\mathcal{Q} = \mathfrak{P}
\left(\mathcal{U}\mathcal{Z}
+ \mathfrak{C}_{=2}(\mathcal{Z})
+ \mathfrak{C}_{=3}(\mathcal{Z})
+ \mathfrak{C}_{=4}(\mathcal{Z})
+ \cdots\right).$$
Translating to generating functions we obtain
$$Q(z, u) = \exp\left(
uz + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \cdots \right)
\\= \exp\left(uz -z + \log\frac{1}{1-z}\right)
= \frac{e^{-z}}{1-z} e^{uz}.$$
The probability that no one gets their hat is computed from the count
of permutations with no fixed points,
$$D_n = n! [z^n] [u^0] Q(z, u)
= n! [z^n] \frac{e^{-z}}{1-z}
= n! \sum_{k=0}^n \frac{(-1)^k}{k!} \approx \frac{n!}{e}.$$
It follows that the probability is $1/e.$
For the expected value of people getting their hat we obtain
$$[z^n] \left. \frac{d}{du} Q(z, u)\right|_{u=1}
= [z^n] \left. \frac{z e^{-z}}{1-z} e^{uz}\right|_{u=1}
= [z^n] \frac{z}{1-z} = 1$$
and thus there is an average of one fixed point per permutation.
As for the location of the smallest fixed point call it $q$ we can
reason as follows. Take any permutation $\sigma$ of the $n-q$ elements
that are larger than $q$ and factor it into cycles. Partition the
$q-1$ elements smaller than $q$ into a derangement containing $p$
elements and distribute the remaining $q-1-p$ elements onto the cycles
of the permutation $\sigma$ of $n-q$, which will assure that none of
them becomes a fixed point and $q$ is the smallest fixed point because
the elements less than $q$ are either in the derangement of $p$ elements
or on the cycles of $\sigma,$ where none of them is a singleton.
This gives for $p$ fixed the value
$$(n-q)! \times {q-1\choose p} \times D_p \times
(n-q)(n-q+1)(n-q+2)\\ \cdots (n-q+(q-1-p-1))
\\= (n-q)! \times {q-1\choose p} \times D_p \times
(n-q)(n-q+1)(n-q+2)\ldots (n-p-2)
\\= (n-q)! \times {q-1\choose p} \times D_p \times
{n-p-2\choose q-1-p} (q-1-p)!$$
When $q<n$ this can also be written as
$$(n-q)! \times {q-1\choose p} \times D_p \times
\frac{(n-p-2)!}{(n-q-1)!}$$
or
$$(n-q) \times {q-1\choose p} \times D_p \times (n-p-2)!$$
The fourth term in the product above on the first line i.e. the
product at the end represents the fact that we have $n-q$ choices when
placing the first element from the $q-1-p$ elements on the cycles of
$\sigma,$ and $n-q+1$ choices for the next one and so on.
The first binomial coeffcient represents the choice of $p$ elements for
the derangement.
The total contribution which is the desired sum of the smallest fixed
point over all permutations is thus given by
$$\sum_{q=1}^n q\times (n-q)! \times
\sum_{p=0}^{q-1} {q-1\choose p} \times D_p \times
{n-p-2\choose q-1-p} (q-1-p)!$$
or alternatively
$$n\times D_{n-1} + \sum_{q=1}^{n-1} q\times (n-q)\times
\sum_{p=0}^{q-1} {q-1\choose p} \times D_p \times (n-p-2)!$$
where we have used the fact that if $n$ is the first fixed point it must have been preceded by a derangement of the first $n-1$ elements.
This yields the sequence
$$1, 1, 7, 31, 191, 1331, 10655, 95887, 958879, 10547659, \ldots$$
which points us to OEIS A155521 where
additional material awaits and which would appear to confirm the above
calculation.
The following admittedly simple Maple program was used to confirm the
above formulae for small $n:$
P := proc(n)
local gf, p, pos;
option remember;
gf := 0;
for p in combinat:-permute(n) do
for pos to n do if p[pos] = pos then break end if end do;
if pos <= n then gf := gf + z^pos end if
end do;
gf
end proc
Addendum.
(Inspired by the work of @YuvalFilmus.)
We learn from the OEIS entry that by a combinatorial bijection the
sequence above is the number of permutations on $n+1$ elements having
at least two fixed points. This has EGF
$$Q_{\ge 2}(z) =
\left. (Q(z, u) - [u^0] Q(z, u) - [u^1] Q(z, u))\right|_{u=1}
= \frac{1}{1-z}
- \frac{e^{-z}}{1-z}
- \frac{ze^{-z}}{1-z}.$$
Extracting coefficients we get
$$n! [z^n] Q_{\ge 2}(z) =
n! \left(1 - \sum_{k=0}^n \frac{(-1)^k}{k!}
- \sum_{k=0}^{n-1} \frac{(-1)^k}{k!}\right)
\approx n! \left(1-\frac{2}{e}\right)$$
Dividing by the count of non-derangements we obtain
$$\frac{(n+1)!(1-2/e)}{n!(1-1/e)} =
(n+1) \times \frac{e-2}{e-1}.$$
Best Answer
Let $E_i$ be the event that professor $i$ gets their hat back. First find $P(E_1\cap E_2^c\cap E_3^c\cap \dots\cap E_n^c)$, the probability that the first professor gets their hat back and no one else does. The final answer is $n$ times this probability. $$ \begin{align} P(E_1\cap E_2^c\cap E_3^c\cap \dots\cap E_n^c) =P(E_2^c\cap \dots \cap E_n^c|E_1)P(E_1) \end{align} $$ Now, given that $E_1$ has occurred, we are just down to the $n-1$ professor situation, so $$ P(E_2^c\cap \dots \cap E_n^c|E_1)= \frac1{2!}-\frac1{3!}+\dots +\frac{(-1)^{n-1}}{(n-1)!} $$ Putting this all together gives you your answer.