= Solution
Use <cycles-per-time spectral density>, with frequency $\omega\in[-1/2,1/2]$. The <spectral representation theorem for a stationary time series> gives the centered <white noise> representation
$$
\varepsilon_t=\int_{-1/2}^{1/2}e^{2\pi it\omega}\,dZ_\varepsilon(\omega),\qquad
\mathbb E|dZ_\varepsilon(\omega)|^2=\sigma^2\,d\omega,
$$
where disjoint increments are orthogonal. Put $\theta_0=1$. Since the filter is finite, substitute each noise representation and interchange the finite sum with the integral:
$$
X_t=\int_{-1/2}^{1/2}e^{2\pi it\omega}\Theta(e^{-2\pi i\omega})\,dZ_\varepsilon(\omega).
$$
The new orthogonal increment measure is $dZ_X=\Theta(e^{-2\pi i\omega})dZ_\varepsilon$. Its <variance> measure is therefore $\sigma^2|\Theta(e^{-2\pi i\omega})|^2d\omega$. The coefficients are real, so conjugation changes the sign of the exponent without changing the modulus. Hence
$$
\boxed{f_X(\omega)=\sigma^2|\Theta(e^{2\pi i\omega})|^2.}
$$
With angular frequency $\lambda=2\pi\omega$, the <spectral density of a stationary process> instead contains the factor $1/(2\pi)$. The convention explains its absence here. Invertibility is not needed for this finite-filter spectral calculation.
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