Solution:

a.
If $m$ is the smallest positive integer such that $f(m) < m$, the $m-1$ numbers preceding $m$ must all have image smaller than $f(m)$ and therefore cannot all be different, since at most $m-2$ values are available, a contradiction.

b.
If $m$ is the smallest positive integer such that $f(m) > m$ and is even, then $f(m) = 2 f\left(\frac{m}{2}\right) = 2 \frac{m}{2} = m$, a contradiction.
If $m$ is odd, then $f(m+1) = 2 f\left(\frac{m+1}{2}\right) = m+1$ since $\frac{m+1}{2} < m$, because $m > 2$, hence $f(m) < m+1$, a contradiction.

c.
Yes. Indeed, if $m$ is the smallest positive integer such that $f(m) < m$ and is even, then one proceeds as before if $\frac{m}{2}$ is odd, and one passes to $f(m+2) = 2 f\left(\frac{m+2}{2}\right) = m+2$ if $\frac{m}{2}$ is even (note that $\frac{m+2}{2} < m$ because $m > 2$).
If $m$ is odd, again one proceeds as before if $\frac{m+1}{2}$ is odd, otherwise one passes to $f(m+3) = 2 f\left(\frac{m+3}{2}\right) = m+3$ because $\frac{m+3}{2} < m$, obviously provided that $m > 3$. It thus remains to exclude the case $m = 3$, that is, one must prove that $f(3) = 3$. To this end let $d$ be an odd number, not divisible by $3$, and such that $\frac{3d+1}{2}$ is odd (for example $d = 7$ works). Then we have:
$$\begin{gathered} f\left(\frac{3d+1}{2}\right) < f(2d) = 2 f(d) \\ f(3d+1) = 2 f\left(\frac{3d+1}{2}\right) > f(3d) = f(3) f(d) \end{gathered}$$
from which $2 > \frac{f(3)}{2}$ and hence $f(3) < 4$.
We observe that in a more direct way one can prove that $f(3) = 3$ as follows.
Let us set $f(3) = x$.
By monotonicity we have $f(5) \geq x+2$ and, by multiplicativity, $f(15) \geq x(x+2) = x^{2} + 2x$. Using alternately multiplicativity and monotonicity we also have:
$$\begin{aligned} & f(6) = 2x \\ & f(5) \leq 2x-1 \\ & f(10) \leq 4x-2 \\ & f(9) \leq 4x-3 \\ & f(18) \leq 8x-6 \end{aligned}$$
Again by monotonicity, we have $f(15) + 3 \leq f(18)$, from which
$$\begin{gathered} x^{2} + 2x + 3 \leq 8x - 6 \\ x^{2} - 6x + 9 = (x-3)^{2} \leq 0 \end{gathered}$$
and hence $x = 3$.