The correct answer is $\boxed{\text{B) }1}$.
The exponential Fourier series of a signal $x(t)$ is given by
$$x(t) = \sum_{k=-\infty}^{\infty} C_k e^{j\omega_k t}$$
where $\omega_k = 2\pi k/T$ is the $k$th harmonic frequency and $T$ is the period of the signal. The coefficients $C_k$ are given by
$$C_k = \frac{1}{T} \int_{0}^{T} x(t) e^{-j\omega_k t} dt$$
In this case, the input signal is $x(t) = 2\cos(2\pi t/3) – \cos(\pi t)$. The Fourier transform of this signal is
$$X(j\omega) = \frac{1}{2} \left(e^{j\pi/3} + e^{-j\pi/3} – 1\right)$$
The transfer function of the system is $H(s) = e^{s} + e^{-s}$. The Fourier transform of the output signal is
$$Y(j\omega) = H(j\omega) X(j\omega) = \frac{1}{2} \left(e^{j\pi/3} + e^{-j\pi/3} – 1\right) \left(e^{j\omega} + e^{-j\omega}\right)$$
The coefficients of the exponential Fourier series of the output signal are given by
$$C_k = \frac{1}{T} \int_{0}^{T} Y(j\omega) e^{-j\omega_k t} dt = \frac{1}{2} \left(e^{j\pi/3} + e^{-j\pi/3} – 1\right) \left(\frac{1}{T} \int_{0}^{T} e^{j(\omega – \omega_k) t} dt + \frac{1}{T} \int_{0}^{T} e^{j(\omega + \omega_k) t} dt\right)$$
The first integral is zero because the integrand is an odd function. The second integral is
$$\frac{1}{T} \int_{0}^{T} e^{j(\omega + \omega_k) t} dt = \frac{e^{j\omega_k T} – 1}{j\omega_k}$$
Therefore, the coefficients of the exponential Fourier series of the output signal are
$$C_k = \frac{1}{2} \left(e^{j\pi/3} + e^{-j\pi/3} – 1\right) \left(\frac{e^{j\omega_k T} – 1}{j\omega_k} + \frac{e^{-j\omega_k T} – 1}{-j\omega_k}\right)$$
For $k = 3$, this is
$$C_3 = \frac{1}{2} \left(e^{j\pi/3} + e^{-j\pi/3} – 1\right) \left(\frac{e^{j3\pi/3} – 1}{j3\pi/3} + \frac{e^{-j3\pi/3} – 1}{-j3\pi/3}\right) = 1$$