(a)
in the polar variables:\\
$$ \Delta u(r,\theta,\phi) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}(u_{\theta\theta} +\cot(\theta)u_\theta + \frac{1}{\sin(\theta)^2}u_{\phi\phi}) = k^2 u $$
Since we are looking for $u(r,\theta,\phi)=u(r)$ . Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, so we have:
$$ \Delta u(r) := u_{rr} + \frac{2}{r} u_{r} = k^2 u $$
Substituting in $ u(r) = \frac{v(r)}{r} $ then:
$$u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $$,
$$u_{rr} = \frac{v_{rr}}{r} - 2\frac{v_{r}}{r^2}+ 2\frac{v}{r^3} $$
$$\implies$$
$$ \Delta u := (\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}) + \frac{2}{r} ( \frac{v_r}{r} - \frac{v}{r^2} ) = k^2 \frac{v}{r} $$
$$\implies$$
$$ \frac{v_{rr}}{r}=k^2\frac{v}{r}\implies v_{rr} = k^2 v $$
Since we have that $ k >0 $, $k^2 >0$, :
$$ v(r) = c_1 e^{k r} + c_2 e^{-k r} = c_3 \cosh(k r) + c_4 \sinh(k r) $$
$$ \implies u(r) = \frac{1}{r} v(r) = c_1\frac{e^{k r}}{r}+c_2\frac{e^{-k r}}{r} =c_3\frac{\cosh(k r)}{r} + c_4 \frac{\sinh(k r)}{r}\phantom{\ } $$
where $ \{c_1,c_2,c_3,c_4\} \in \mathbb{R} $
(b)
Similar to a:\\
$$ \Delta u(r,\theta,\phi) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}(u_{\theta\theta} +\cot(\theta)u_\theta + \frac{1}{\sin(\theta)^2}u_{\phi\phi}) = -k^2 u $$
Since we are looking for $u(r,\theta,\phi)=u(r)$ . Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, so we have:
$$ \Delta u(r) := u_{rr} + \frac{2}{r} u_{r} = -k^2 u $$
Substituting in $ u(r) = \frac{v(r)}{r} $ then:
$$u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $$,
$$u_{rr} = \frac{v_{rr}}{r} - 2\frac{v_{r}}{r^2}+ 2\frac{v}{r^3} $$
$$\implies$$
$$ \Delta u := (\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}) + \frac{2}{r} ( \frac{v_r}{r} - \frac{v}{r^2} ) = -k^2 \frac{v}{r} $$
$$\implies$$
$$ v_{rr} = - k^2 v $$
Since we have that $ k >0 $, $k^2 >0$, :
$$ v(r) = c_1 \sin (k r) + c_2 \cos (k r) $$
$$ \implies u(r) = \frac{v}{r} = c_1\frac{\sin (k r)}{r}+c_2\frac{\cos (k r)}{r} $$
where $ \{c_1,c_2\} \in \mathbb{R} $
