(a)
The form of Bessel's function,
$$\frac {\partial^2 y}{\partial x^2}+\frac{1}{x} \frac{\partial y}{\partial x}+(1+\frac{n^2}{x^2})y=0$$
we can rewrite the equation:
$$ u_{rr} + \frac{1}{r} u_{r} +(1-\frac{s^2}{r^2})u = 0 $$
$$ \text{Let: } u(r) = v(i k r) \implies u_{r} = i k v_{r}, \phantom{\ } u_{rr} = - k^2 v_{rr} $$
$$ \implies \Delta_{(r,\theta)} u(r) := u_{rr} + \frac{1}{r} u_{r} - k^2 u = 0$$
$$- k^2 v_{rr} + \frac{1}{r} i k v_{r} - k^2 v = 0 $$
Since $ k >0$, $ k^2 \ne 0 $, so:
$$ v_{rr} - \frac{i}{k r} v_{r} + v = 0 $$
$ - \frac{i}{k r} = \frac{(-i)}{(1) k r} = \frac{(-i)}{(-i * i) k r} = \frac{\overline{i}}{\overline{i}} \frac{1}{i k r} = \frac{1}{i k r} $ so our ODE in $v(i k r)$, say $v(z)$ where $z = i k r$ is equivalent to:
$$ v_{rr} + \frac{1}{i k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$
Since $v_{rr} + \frac{1}{i k r} v_{r} = -v $ is a linear combination of Bessel funtion of $J_0 \& N_0$
$$ \implies v(z) = v(i k r) = u(r) = A_0 J_0(i k r) + B_0 N_0(i k r) \phantom{\ } $$
(b)
As (a):
$$ \Delta_{(r,\theta)} u(r,\theta) := u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^2} u_{\theta\theta} = u_{rr} + \frac{1}{r} u_{r} = - k^2 u $$
rewrite in Bessel equation:
$$ u_{zz} + \frac{1}{z} u_{z} +(1-\frac{s^2}{z^2})u = 0 $$
$$ \text{Let: } u(r) = v(k r) \implies u_{r} = k v_{r}, \phantom{\ } u_{rr} = k^2 v_{rr} $$
$$ \implies \Delta_{(r,\theta)} u(r) := u_{rr} + \frac{1}{r} u_{r} + k^2 u = 0 arrow k^2 v_{rr} + \frac{1}{r} k v_{r} + k^2 v = 0 $$
Let $z = k r$. Again, $k >0$ so $k^2 \ne 0$ :
$$ v_{rr} + \frac{1}{k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$
Which is clearly a Bessel's differential equation with $s = n = 0$:
$$ v_{zz} + \frac{1}{z} v_{z} + v = v_{zz} + \frac{1}{z} v_{z} +(1-\frac{s^2}{z^2})v = 0 $$
So our solution for $v(z)$ is again $ v(z) = A J_0(z) + B N_0(z) $ for some $ \{A,B\} \in \mathbb{R} $.
$$ \implies v(z) = v(k r) = u(r) = A_0 J_0(k r) + B_0 N_0(k r) \phantom{\ } $$
