To prove 2), it is convenient to show that such a function, if it has only finitely many discontinuities, must be discontinuous at $1$.
So let $f$ be such a function, and suppose it has only finitely many discontinuities. So we have a finite partition
$$\frac{1}{a} = t_0 < t_1 < \dotsc < t_{n-1} < t_n = a$$
of the interval, so that $f$ is continuous on each open interval $I_K := (t_k, t_{k+1})$, $0 \leqslant k < n$, and $f$ is discontinuous at $t_k$ for $0 < k < n$. Whether $f$ is continuous at $t_0$ or $t_n$ isn't of interest now. We want to show that $t_k = 1$ for some $0 < k < n$.
For each $0 \leqslant k < n$, the restriction $f_k = f\lvert_{I_k}$ is a continuous injective function, hence it is strictly monotonic. And $f_k(I_k) = f(I_k)$ is an open interval on which $f^{-1}$ is continuous. Since $f^{-1} = \frac{1}{f}$, $f$ is continuous at $x$ if and only if $f^{-1}$ is continuous at $x$, so there is an $0 \leqslant \ell < n$ with $f(I_k) \subset I_\ell$. We can apply the same argument to $f^{-1}$ and $I_\ell$ to see that $f^{-1}(I_\ell) \subset I_m$ for some $0 \leqslant m < n$. But
$$I_k = f^{-1}(f(I_k)) \subset f^{-1}(I_\ell) \subset I_m$$
implies $m = k$, and thus $f(I_k) = I_\ell$. If $f$ is increasing (decreasing) on $I_\ell$, then $\frac{1}{f} = f^{-1}$ is decreasing (increasing) on $I_\ell$. But $f^{-1}\lvert_{I_\ell} = (f_k)^{-1}$, so then $f_k^{-1}$ and therefore $f_k$ is decreasing (increasing). Thus none of the $I_k$ can be mapped to itself.
Now suppose $f$ were continuous at $1$, say $1 \in I_\ell$. Let $I_k = f^{-1}(I_\ell)$. Since $f$ attains values less than $1$ and values greater than $1$ on $I_k$, so does $\frac{1}{f} = f^{-1}$. Therefore we have $f(I_k) = I_\ell = f^{-1}(I_k)$, and hence $f(I_\ell) = I_k = f^{-1}(I_\ell)$. But if $f(I_\ell) = I_k \subset (1,a)$, then $f^{-1}(I_\ell) \subset (1/a,1)$, so $f^{-1}(I_\ell) \neq f(I_\ell)$ and if $f(I_\ell) \subset (1/a,1)$, then $f^{-1}(I_\ell) \subset (1,a)$, which again implies $f(I_\ell) \neq f^{-1}(I_\ell)$, and we've reached a contradiction.
For part 3), the above considerations show that we need at least four intervals of continuity. Let's try with the partition
$$\frac{1}{a} = t_0 < t_1 < t_2 = 1 < t_3 < t_4 = a.$$
If $f(I_k) = I_1$, then we must have $f^{-1}(I_k) = I_2$, and that shows that necessarily $t_1 = t_3^{-1}$, and we can have either the cycle $I_0 \to I_1 \to I_3 \to I_2 \to I_0$ or $I_0 \to I_2 \to I_3 \to I_1 \to I_0$. Let's take $t_3 = \sqrt{a}$, and the first cycle and see where we get. For the map $I_0 \to I_1$, we can try the straightforward $x \mapsto \sqrt{a}\cdot x$, and from this we find $f^{-1}(x) = \frac{x}{\sqrt{a}}$, so $f(x) = \frac{\sqrt{a}}{x}$ on $I_1$. This forces $f(x) = \frac{1}{\sqrt{a}\cdot x}$ on $I_2$, therefore $f^{-1}(x) = \sqrt{a}\cdot x$ on $I_2$, and that means $f(x) = \frac{x}{\sqrt{a}}$ on $I_3$. We can extend this to one of the boundary points of the respective intervals, and thus obtain
$$f(x) = \begin{cases} \sqrt{a}\cdot x &, \frac{1}{a} \leqslant x < \frac{1}{\sqrt{a}} \\ \frac{\sqrt{a}}{x} &, \frac{1}{\sqrt{a}} \leqslant x < 1 \\ 1 &, x = 1 \\ \frac{1}{\sqrt{a}\cdot x} &, 1 < x \leqslant \sqrt{a} \\ \frac{x}{\sqrt{a}} &, \sqrt{a} < x \leqslant a.\end{cases}$$
This function satisfies the requirements.
