Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the inductor and the resistor are in series, they have the same current.
$$I_R=I_L=I$$
For the resistor, we know that $I_R$ has the same direction of $V_R$ while for the inductor, $I_L$ lags the $V_L$ by an angle of $90^\circ$.
See the figure below.
According to Ohm's law,
$$I=\dfrac{\varepsilon_0}{\sqrt{X_L^2+R^2}}$$
where $X_L=\omega L=2\pi fL$
$$\boxed{I=\dfrac{\varepsilon_0}{\sqrt{\omega^2L^2+R^2}}}\tag 1$$
Recalling that
$$V_L=IX_L=I\omega L$$
Plug from (1),
$$\boxed{V_L= \dfrac{\varepsilon_0\omega L}{\sqrt{\omega^2L^2+R^2}}}$$
And that
$$V_R=IR $$
Plug from (1),
$$\boxed{V_R= \dfrac{R\varepsilon_0}{\sqrt{\omega^2L^2+R^2}}}$$
$$\color{blue}{\bf [b]}$$
when $\omega$ approaches zero, then
$$V_R= \dfrac{R\varepsilon_0}{\sqrt{0+R^2}}=\boxed{\varepsilon_0}$$
and when $\omega$ approaches infinity, then
$$V_R= \dfrac{R\varepsilon_0}{\sqrt{\infty^2+R^2}}=\boxed{0\;\rm V}$$
$$\color{blue}{\bf [c]}$$
If the output is taken from the resistor, it is a low-pass filter $RL$ circuit. That's because when $\omega\rightarrow 0$, the $V_R $ is at its maximum value, and when $\omega\rightarrow \infty$, the $V_R$ is at its lowest value.
$$\color{blue}{\bf [d]}$$
The cross-over frequency occurs at $V_L=V_R$, so
$$IX_L=IR$$
So,
$$\omega_c L=R$$
and hence,
$$\boxed{\omega_c =\dfrac{R}{L}}$$
