Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
As we see below, the inductor voltages lead the current by 90$^\circ$, while the capacitor voltage lags the current by 90$^\circ$. The resistor voltage and current are in phase.
The capacitor and the inductor are in parallel, so
$$V_L=V_C=V_{LC}\tag 1$$
The compensation of the in-parallel capacitor and the inductor is in series to the resistor, so
$$I_{LC}=I_R=I_L-I_C=I\tag 2$$
Also, we can see, from the figure below, that
$$\varepsilon_0^2=V_R^2+V_{LC}^2 \tag 3$$
From (2),
$$I =\dfrac{V_L}{X_L}-\dfrac{V_C}{X_C}=\dfrac{V_{LC}}{X_L}-\dfrac{V_{LC}}{X_C}$$
$$I = V_{LC}\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]$$
So,
$$V_C=\dfrac{I}{\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]}$$
Plug into (3),
$$\varepsilon_0^2=V_R^2+\dfrac{I^2}{\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]^2}$$
$$\varepsilon_0^2=I^2R^2+\dfrac{I^2}{\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]^2}$$
Solving for $I$;
$$\boxed{I=\dfrac{\varepsilon_0}{\sqrt{R^2+\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]^{-2}}}}$$
$$\color{blue}{\bf [b]}$$
When $\omega\rightarrow0$, then $X_L\rightarrow0$, and $X_C\rightarrow \infty$, and hence, $\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]^{-2}=\left[\infty\right]^{-2}\rightarrow 0$
Thus,
$$\boxed{I=\dfrac{\varepsilon_0}{R}}$$
And when $\omega\rightarrow\infty$, then $X_L\rightarrow \infty$, and $X_C\rightarrow 0$, and hence, $\left[\dfrac{1}{X_L}-\dfrac{1}{X_C}\right]^{-2}=\left[-\infty\right]^{-2}\rightarrow 0$
Thus,
$$\boxed{I=\dfrac{\varepsilon_0}{R}}$$
$$\color{blue}{\bf [c]}$$
The resonance frequency occurs when $X_C=X_L$ and this here means that $I_C=I_L$,
So, from (2),
$$\boxed{I=0\;\rm A}$$
