Solution

a) The field inside the capacitor is $E$. Each of the two plates individually creates a field $E_0 = E/2$. The plates are oppositely charged, so their emitted fields have opposite signs.
Let the plate closer to the observer be at distance $x$, and the farther one at $x+d$. According to the formula, the total radiated field is:
$$E_{\text{rad}} = \frac{1}{c} \left( v_{t - \frac{x+d}{c}} \frac{E}{2} - v_{t - \frac{x}{c}} \frac{E}{2} \right) = \frac{E}{2c} \left( v\left(t - \frac{x+d}{c}\right) - v\left(t - \frac{x}{c}\right) \right)$$
Since the capacitor moves with constant acceleration $a$, then $v(t) = at$. The difference in velocities for times differing by $\Delta t = d/c$ is $\Delta v = a \frac{d}{c}$.
Then the magnitude of the radiated field is:
$$E_{\text{rad}} = \frac{E}{2c} \left( a \frac{d}{c} \right) = \frac{a d E}{2 c^2}$$

b) The surface current $i = i_0 \sin(\omega t)$ creates a magnetic field $B$ near the plate.
In SI: By Ampere's law, $B = \frac{\mu_0 i}{2}$. The emitted electric field is related to the magnetic field as $E_{\text{rad}} = c B = \frac{c \mu_0 i_0}{2}$. Since $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, the amplitude is:
$$E_{\text{rad}} = \frac{i_0}{2 \varepsilon_0 c}$$
In CGS: The magnetic field of the current is $B = \frac{2\pi i}{c}$. The emitted electric field is $E_{\text{rad}} = B$, hence the amplitude is:
$$E_{\text{rad}} = \frac{2\pi i_0}{c}$$

c) The incident wave $E_{\text{inc}} = E_0 \cos(2\pi\nu t)$ forces the film's electrons to move. The equation of motion for an electron is $m_e \dot{v} = e E_0 \cos(2\pi\nu t)$.
Electron velocity: $v(t) = \frac{e E_0}{m_e \cdot 2\pi\nu} \sin(2\pi\nu t)$.
The current density in the film is $j = n_e e v$, and the surface current (current per unit width) is $K = j x$:
$$K(t) = \frac{n_e e^2 x E_0}{2\pi m_e \nu} \sin(2\pi\nu t)$$
Amplitude of the surface current $K_0 = \frac{n_e e^2 x E_0}{2\pi m_e \nu}$.
The film acts as an emitting plate from part (b). The amplitude of the reflected wave (in SI) is:
$$E_{\text{ref}} = \frac{K_0}{2 \varepsilon_0 c} = \frac{n_e e^2 x E_0}{4\pi \varepsilon_0 c m_e \nu}$$
The reflection coefficient $R$ is the ratio of the intensities (squares of the field amplitudes) of the reflected and incident waves:
$$R = \left( \frac{E_{\text{ref}}}{E_0} \right)^2 = \left( \frac{n_e e^2 x}{4 \pi \varepsilon_0 m_e c \nu} \right)^2$$