Answer
See the detailed answer below.
Work Step by Step
We know that the impedance of the $RLC$-circuit is given by
$$Z=\sqrt{R^2+\left[X_L-X_C\right]}$$
$$Z=\sqrt{R^2+\left[2\pi f L-\dfrac{1}{2\pi f C}\right]^2}\tag 1$$
Hence, the peak current is given by
$$I=\dfrac{\varepsilon_0}{Z}\tag 2$$
And the phase angle is then,
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]$$
$$\phi=\tan^{-1}\left[\dfrac{2\pi f L-\dfrac{1}{2\pi f C}}{R}\right]\tag 3$$
Using (1), (2), and (3) above to find the impedance, the peak current, and the phase angle, respectively, for each given frequency.
$$\color{blue}{\bf [a]}$$
At $f=3000$ Hz,
(1):
$$Z=\sqrt{50^2+\left[2\pi (3000) (3.3\times 10^{-3})-\dfrac{1}{2\pi (3000) (480\times 10^{-9})}\right]^2} $$
$$Z=\color{red}{\bf 69.53}\;\rm \Omega$$
(2):
$$I=\dfrac{5}{69.53 }=\color{red}{\bf 71.9}\;\rm mA $$
(3):
$$\phi=\tan^{-1}\left[\dfrac{2\pi (3000) (3.3\times 10^{-3})-\dfrac{1}{2\pi (3000) (480\times 10^{-9})}}{50}\right] $$
$$\phi=\color{red}{\bf -44.02}^\circ$$
$$\color{blue}{\bf [b]}$$
At $f=4000$ Hz,
(1):
$$Z=\sqrt{50^2+\left[2\pi (4000) (3.3\times 10^{-3})-\dfrac{1}{2\pi (4000) (480\times 10^{-9})}\right]^2} $$
$$Z=\color{red}{\bf 50}\;\rm \Omega$$
(2):
$$I=\dfrac{5}{50}=\color{red}{\bf 100}\;\rm mA $$
(3):
$$\phi=\tan^{-1}\left[\dfrac{2\pi (4000) (3.3\times 10^{-3})-\dfrac{1}{2\pi (4000) (480\times 10^{-9})}}{50}\right] $$
$$\phi=\color{red}{\bf 0}^\circ$$
$$\color{blue}{\bf [c]}$$
At $f=5000$ Hz,
(1):
$$Z=\sqrt{50^2+\left[2\pi (5000) (3.3\times 10^{-3})-\dfrac{1}{2\pi (5000) (480\times 10^{-9})}\right]^2} $$
$$Z=\color{red}{\bf 62.415}\;\rm \Omega$$
(2):
$$I=\dfrac{5}{62.415}=\color{red}{\bf 80.1}\;\rm mA $$
(3):
$$\phi=\tan^{-1}\left[\dfrac{2\pi (5000) (3.3\times 10^{-3})-\dfrac{1}{2\pi (5000) (480\times 10^{-9})}}{50}\right] $$
$$\phi=\color{red}{\bf 36.8}^\circ$$
