Answer
a) ${\bf 31.8}\;\rm kHz$
b) ${\bf 0}\;\rm V$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$V_L=I_LX_L$$
where $X_L=\omega L=2\pi f L$, so
$$V_L=I_LX_L=2\pi f L I_L$$
Solving for $f$;
$$ f =\dfrac{V_L}{2\pi L I_L}$$
Plug the given;
$$ f =\dfrac{5}{2\pi (500\times 10^{-6}) (50\times 10^{-3})}$$
$$f=\color{red}{\bf 31.8}\;\rm kHz$$
$$\color{blue}{\bf [b]}$$
The inductor voltage is given by
$$v_L=V_L\cos(\omega t)=5\cos(\omega t)\tag 1$$
We know that the inductor current lags the inductor voltage by $\pi/2=90^\circ$, so
$$i_L=I_L\cos\left(\omega t-\frac{\pi}{2}\right) $$
So when $i_L=I_L$,
$$\cos\left(\omega t-\frac{\pi}{2}\right)=1$$
which means that
$$\omega t-\frac{\pi}{2}=0$$
and hence,
$$\omega t=\frac{\pi}{2}$$
Plug into (1),
$$v_L=5\cos\left(\frac{\pi}{2}\right)=\color{red}{\bf 0}\;\rm V$$
