(a)The phase difference between current and voltage $\phi$ is given by
$$
\begin{array}{l}
\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=500 \Omega \\
\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=100 \Omega \\
\mathrm{R}=400 \Omega
\end{array}
$$
Thus $\tan \phi=\frac{400}{400}=1$
$\Longrightarrow \phi=45^{\circ}$
(b) Power factor is given by $\cos \phi=1$
$$
\begin{array}{l}
\Longrightarrow \phi=0^{\circ} \\
\Longrightarrow \tan \phi=0 \\
\Longrightarrow \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}=100 \Omega \\
\Longrightarrow \mathrm{C}=10 \mu \mathrm{F}
\end{array}
$$
Hence an $8 \mu \mathrm{F}$ capacitor need to be connected in parallel with the given capacitor.