To find what values $f(7)$ might take, suppose that there are $t$ tails
and $h$ heads in the first $7$ coin tosses. We have $h + t = 7$. Scott
will have moved $h$ steps to the right and $t$ steps to the left so we will
be standing on $h - t$. Substituting $t = 7 - h$ we find
$$
h - t = h - (7-h) = 2h - 7.
$$
Since $2h$ is an even number, $2h - 7$ is an odd number and so it is not
possible that $f(7) = 2$.

Alternatively, we can reason directly that landing on $2$ after $7$ tosses is
not possible. We know that
$h + t = 7$ after seven tosses. Landing on $2$ after the seven tosses
would give us the additional equation $h - t = 2$. Solving these two
equations in two unknowns gives $h = \frac{9}{2}$ and $t = \frac{5}{2}$
which is not possible since the number of heads (and also the number of tails
tails) must be a whole number.

Since Scott is taking a total of $50$ steps, he can not pass $+50$ to the
right or $-50$ to the left. Between $+50$ and $-50$ the argument from
part (b) will show that only even numbers are possible. To see this, if
$h$ is the number of heads and $t$ the number of tails then Scott will
be standing on $h - t$ after the $50$ tosses. We have $h + t = 50$ since
there are $50$ tosses. Substituting $t = 50 - h$ shows that Scott will
be standing on $h - (50-h) = 2h - 50$ after the $50$ coin tosses. Since
$2h$ and $-50$ are even numbers this shows that Scott is on an even number.

So far, we have shown that Scott must end on an even number between $-50$
and $50$.
We have seen that with $h$ heads Scott ends up on $2h - 50$. When
$h$ takes the values $0,1, \ldots, 50$, $2h-50$ takes the values
$$
-50, -48, \ldots, 50
$$
or, in other words, all even numbers between $-50$ and $50$.
So $P(m) = 0$ for any odd integer $m$ and also any even integer $m$
whose absolute value is larger than $50$.