Answer
See the detailed answer below.
Work Step by Step
We know that the instantaneous current is given by
$$i=I\cos(\omega t-\phi) \tag 1$$
where $I=\varepsilon_0/Z$, where $Z=\sqrt{(X_L-X_C)^2+R^2}$
$$i=\dfrac{\varepsilon_0\cos(\omega t-\phi) }{\sqrt{(X_L-X_C)^2+R^2}}$$
where $\phi$ is given by
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]$$
where $X_L=2\pi fL$, and $X_C=1/2\pi f C$
$$\phi=\tan^{-1}\left[\dfrac{2\pi f L-\dfrac{1}{2\pi f C}}{R}\right]$$
Plug the known;
$$\phi=\tan^{-1}\left[\dfrac{2\pi(3\times 10^3) (3.3\times 10^{-3})-\dfrac{1}{2\pi (3\times 10^3)(480\times 10^{-9})}}{(50)}\right]$$
$$\phi=\bf -44.02^\circ$$
Plug into (1),
$$i=I\cos(\omega t+44.02^\circ) \tag 2$$
Recalling that
$$\varepsilon=\varepsilon_0\cos(\omega t)\tag 3$$
$$\color{blue}{\bf [a]}$$
$\bullet$ When $i=I$, then from (2), $\cos(\omega t+44.02^\circ)=1$ which means that $\omega t+44.02^\circ=0^\circ$. Hence, $\omega t=-44.02^\circ$
Plug into (3),
$$\varepsilon=\varepsilon_0\cos(-44.02^\circ)=5\cos(-44.02^\circ)$$
$$\varepsilon=\color{red}{\bf 3.6}\;\rm V$$
$$\color{blue}{\bf [b]}$$
$\bullet$ When $i=0$ and decreasing, then from (2), $\cos(\omega t+44.02^\circ)=0$ which means that $\omega t+44.02^\circ=90^\circ$. Hence, $\omega t=45.98^\circ$
Plug into (3),
$$\varepsilon=\varepsilon_0\cos(45.98^\circ)=5\cos(45.98^\circ)$$
$$\varepsilon=\color{red}{\bf 3.47}\;\rm V$$
$$\color{blue}{\bf [c]}$$
$\bullet$ When $i=-I$, then from (2), $\cos(\omega t+44.02^\circ)=-1$ which means that $\omega t+44.02^\circ=180^\circ$. Hence, $\omega t=136^\circ$
Plug into (3),
$$\varepsilon=\varepsilon_0\cos(136^\circ)=5\cos(136^\circ)$$
$$\varepsilon=\color{red}{\bf -3.6}\;\rm V$$
