Solution

$а)$Let us prove that the current in a cylinder magnetized to saturation occurs only on its outer surface:
$$j=\Delta \times M$$ $$i=\hat{n}\times M$$ где $\hat{n}$ -
normal vector to the surface.

Because the conductor is magnetized to saturation the rotor of $M=0$ .
At the ends the vector product vanishes, so there is no current there either. $\Rightarrow$
the current is only at the ends and according to the vector product rule its linear density is equal to $i=M$

$b)$
The difference in induction arises from the difference in solid angles at which current-carrying surfaces are visible. Similar to point a, currents arise only on the outer surface and are absent at the ends and inside the conductors. In the case of an infinite stub, the solid angle is $\Omega = 4\pi$.
In the case of a square $\Omega' = \frac{8\pi}{3}$ therefore the induction is $n=1.5$ times less.

$c)$The solid angle at which the outer surface along which the current flows is visible is equal to $$\Omega = 4\pi - 2\cdot2\pi (1-cos\alpha)=4\pi cos\alpha$$ where $cos\alpha = \frac{1}{\sqrt{4\frac{r^2}{l^2}+1}}$
So:
$$B = \frac{\mu_0M}{\sqrt{1+4\frac{r^2}{l^2}}}$$
$r/l \to 0$:
$$B \to \mu_0 M$$
$l/r \to 0$:
$$B \to \frac{\mu_0 Ml}{2r}$$
$d)$
Let us imagine the field of the past cylinder as a superposition of the fields of the thin cylinder in the center and the rest of it. $$\hat{B}_{thin}+\hat{B}_{rest}=\hat{B}_{full}$$
$$\hat{B}_{rest} = -(\hat{B}_{thin}-\hat{B}_{full})$$
$$B = \mu_0 M(1-\frac{1}{\sqrt{1+4\frac{r^2}{l^2}}})$$
$r/l \to 0$:
$$B \to \frac{2\mu_0 Mr^2}{l^2}$$
$l/r \to 0$:
$$B \to \mu_0 M$$