Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
For an $RC$ circuit, the capacitor voltage is given by
$$V_C=\dfrac{\varepsilon_0X_C}{\sqrt{R^2+X_C^2}}$$
where $X_C=1/\omega C$;
$$V_C=\dfrac{\varepsilon_0}{\omega C}\dfrac{1}{\sqrt{R^2+\dfrac{1}{\omega^2 C^2}}}$$
Squaring both sides;
$$V_C^2=\dfrac{\varepsilon_0^2}{\omega^2 C^2}\dfrac{1}{ R^2+\dfrac{1}{\omega^2 C^2}}$$
At $V_C=\frac{1}{2}\varepsilon_0$;
$$\frac{1}{4} \color{red}{\bf\not} \varepsilon_0^2=\dfrac{ \color{red}{\bf\not} \varepsilon_0^2}{\omega^2 C^2}\dfrac{1}{ R^2+\dfrac{1}{\omega^2 C^2}}$$
$$ 1=\dfrac{ 4}{\omega^2 C^2}\dfrac{1}{ R^2+\dfrac{1}{\omega^2 C^2}}$$
$$ 1= \dfrac{4}{ \omega^2R^2C^2+1}$$
$$\omega^2R^2C^2+1=4$$
$$\omega =\sqrt{\dfrac{3}{R^2C^2}}$$
$$\boxed{\omega =\dfrac{\sqrt{3}}{R C }}$$
$$\color{blue}{\bf [b]}$$
For an $RC$ circuit, the resistor voltage is given by
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+X_C^2}}$$
where $X_C=1/\omega C$;
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+\dfrac{1}{\omega^2 C^2}}}$$
Plugging from the boxed formula above.
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+\dfrac{1}{\dfrac{3}{R^2 \color{red}{\bf\not} C^2} \color{red}{\bf\not} C^2}}}$$
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+\dfrac{R^2}{3 }}} =\dfrac{\varepsilon_0 }{\sqrt{1+\dfrac{1}{3 }}}$$
$$V_R=\dfrac{\sqrt{3}\;\varepsilon_0}{2}$$
