Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The crossover frequency in a simple RC high-pass filter circuit is at $V_R=V_C$, where the crossover frequency is given by
$$\omega_c=2\pi f_c$$
So,
$$f_c=\dfrac{\omega_c}{2\pi}$$
where $\omega_c=1/RC$, so
$$f_c=\dfrac{1}{2\pi RC}$$
Plug the known;
$$f_c=\dfrac{1}{2\pi (100)(1.59\times 10^{-6})}=\color{red}{\bf 1001}\;\rm Hz$$
$$\color{blue}{\bf [b]}$$
Recalling that the voltage resistor in such circuits is given by
$$V_R= \dfrac{\varepsilon_0 R}{\sqrt{R^2+X_C^2}}$$
where $X_C=1/\omega C$
$$V_R= \dfrac{\varepsilon_0 R}{\sqrt{R^2+\left[\dfrac{1}{\omega C}\right]^2}}$$
$$V_R= \dfrac{\varepsilon_0}{\sqrt{1+\left[\dfrac{1}{\omega C R}\right]^2}}$$
where $1/RC=\omega_c$
$$V_R= \dfrac{\varepsilon_0}{\sqrt{1+\left[\dfrac{\omega_c}{\omega }\right]^2}}$$
where $\omega=2\pi f$, and $\varepsilon_0=5$ V. so
$$\boxed{V_R= \dfrac{5}{\sqrt{1+\left[\dfrac{f_c }{f }\right]^2}}}$$
$\bullet\Rightarrow$ when $f=\frac{1}{2}f_c$
$$ V_R= \dfrac{5}{\sqrt{1+\left[\dfrac{ \color{red}{\bf\not} f_c }{\frac{1}{2} \color{red}{\bf\not} f_c}\right]^2}} = \sqrt5$$
$$V_R=\color{red}{\bf 2.24}\;\rm V$$
$\bullet\Rightarrow$ when $f= f_c$
$$ V_R= \dfrac{5}{\sqrt{1+\left[\dfrac{ \color{red}{\bf\not} f_c }{ \color{red}{\bf\not} f_c}\right]^2}}=5/\sqrt2$$
$$V_R=\color{red}{\bf 3.54}\;\rm V$$
$\bullet\Rightarrow$ when $f= 2f_c$
$$ V_R= \dfrac{5}{\sqrt{1+\left[\dfrac{ \color{red}{\bf\not} f_c }{ 2\color{red}{\bf\not} f_c}\right]^2}}= 2\sqrt5$$
$$V_R=\color{red}{\bf 4.47}\;\rm V$$
