Answer
a) ${\bf 64.2}\;\rm mA$
b) ${\bf 48}\;\rm mA$
Work Step by Step
We know that the instantaneous current is given by
$$i=I\cos(\omega t-\phi)\tag1 $$
where $I=\varepsilon_0/Z$, where $Z=\sqrt{(X_L-X_C)^2+R^2}$
$$i=\dfrac{\varepsilon_0\cos(\omega t-\phi) }{\sqrt{(X_L-X_C)^2+R^2}}\tag 1$$
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;
$$\phi=\tan^{-1}\left[\dfrac{2\pi(5\times 10^3) (3.3\times 10^{-3})-\dfrac{1}{2\pi (5\times 10^3)(480\times 10^{-9})}}{(50)}\right]$$
$$\phi=\bf 36.77^\circ$$
Plug into (1),
$$i=\dfrac{\varepsilon_0\cos(\omega t-36.77^\circ) }{\sqrt{\left(2\pi f L-\dfrac{1}{2\pi f C}\right)^2+R^2}}\tag 2$$
Recalling that
$$\varepsilon=\varepsilon_0\cos(\omega t)\tag 3$$
$$\color{blue}{\bf [a]}$$
$\bullet$ When $\varepsilon=\varepsilon_0$, then from (3), $\cos(\omega t)=1$ which means that $\omega t=0$.
Plug into (2),
$$i=\dfrac{\varepsilon_0\cos(0^\circ-36.77^\circ) }{\sqrt{\left(2\pi f L-\dfrac{1}{2\pi f C}\right)^2+R^2}} $$
Plug the known;
$$i=\dfrac{(5)\cos( -36.77^\circ) }{\sqrt{\left(2\pi(5\times 10^3)(3.3\times 10^{-3})-\dfrac{1}{2\pi (5\times 10^3)(480\times 10^{-9})}\right)^2+(50)^2}} $$
$$i =\color{red}{\bf 64.2}\;\rm mA$$
$$\color{blue}{\bf [b]}$$
$\bullet$ When $\varepsilon=0$ and $\varepsilon$ decreases, then from (3), $\cos(\omega t)=0$ which means that $\omega t=90^\circ$.
Plug into (2),
$$i=\dfrac{\varepsilon_0\cos(90^\circ-36.77^\circ) }{\sqrt{\left(2\pi f L-\dfrac{1}{2\pi f C}\right)^2+R^2}} $$
Plug the known;
$$i=\dfrac{(5)\cos(90^\circ -36.77^\circ) }{\sqrt{\left(2\pi(5\times 10^3)(3.3\times 10^{-3})-\dfrac{1}{2\pi (5\times 10^3)(480\times 10^{-9})}\right)^2+(50)^2}} $$
$$i= \color{red}{\bf 48}\;\rm mA$$
