Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
For an $RC$ circuit, the resistor voltage is given by
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+X_C^2}}$$
where $X_C=1/\omega C$;
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+\dfrac{1}{\omega^2 C^2}}}$$
Squaring both sides;
$$V_R^2=\dfrac{\varepsilon_0^2R^2}{ R^2+\dfrac{1}{\omega^2 C^2}}$$
At $V_R=\frac{1}{2}\varepsilon_0$;
$$\frac{1}{4}\color{red}{\bf\not}\varepsilon_0^2=\dfrac{\color{red}{\bf\not}\varepsilon_0^2R^2}{ R^2+\dfrac{1}{\omega^2 C^2}}$$
$$R^2+\dfrac{1}{\omega^2 C^2}= 4R^2 $$
$$ \dfrac{1}{\omega^2 C^2}= 3R^2 $$
$$ \omega =\sqrt{\dfrac{1}{3R^2 C^2} }$$
$$ \boxed{\omega =\dfrac{1}{\sqrt{3}\;R C }}$$
$$\color{blue}{\bf [b]}$$
For an $RC$ circuit, the capacitor voltage is given by
$$V_C=IX_C=\dfrac{V_R}{R}\cdot \dfrac{1}{\omega C}$$
$$V_C= \dfrac{\varepsilon_0}{2R}\cdot \dfrac{1}{\omega C}$$
Plugging from the boxed formula above.
$$V_C= \dfrac{\varepsilon_0}{2\color{red}{\bf\not} R}\cdot \dfrac{\sqrt{3}\;\color{red}{\bf\not} R \color{red}{\bf\not} C}{ \color{red}{\bf\not} C}$$
$$\boxed{V_C= \dfrac{\sqrt{3}\;\varepsilon_0}{2}}$$
