= Solution
Use the two-sided <spectral density of a stationary process> convention
$$
\boxed{\gamma_k=\int_{-\pi}^{\pi}e^{ik\lambda}f_X(\lambda)\,d\lambda.}
$$
Thus <white noise> of <variance> $\sigma^2$ has density $\sigma^2/(2\pi)$.
Absolute summability makes the <linear filter of a stationary time series> well defined in $L^2$: $\sum_r\|a_rX_{t-r}\|_2\leq\|X_0\|_2\sum_r|a_r|<\infty$. Its <mean> is $m_Y=m_X\sum_ra_r$. For its <autocovariance>, continuity of the $L^2$ inner product gives
$$
\gamma_Y(k)=\sum_{r,s}a_ra_s\gamma_X(k-r+s).
$$
The sum is absolutely convergent because $|\gamma_X(j)|\leq\gamma_X(0)$. It depends only on $k$, proving <weak stationarity>. Inserting the spectral inversion formula and exchanging the absolutely dominated sum with the integral yields
$$
\begin{aligned}
\gamma_Y(k)&=\int_{-\pi}^{\pi}e^{ik\lambda}
\left(\sum_ra_re^{-ir\lambda}\right)
\left(\sum_sa_se^{is\lambda}\right)f_X(\lambda)\,d\lambda,\\
\boxed{f_Y(\lambda)}&=\boxed{|a(\lambda)|^2f_X(\lambda)}.
\end{aligned}
$$
Here the coefficients are real, so the two factors are complex conjugates. This is the <spectral density transformation under a linear filter>.
Applying a second <linear filter of a stationary time series> gives
$$
\boxed{f_Z(\lambda)=|b(\lambda)|^2|a(\lambda)|^2f_X(\lambda).}
$$
The <composition of absolutely summable time-series filters> can also be calculated directly:
$$
Z_t=\sum_jb_j\sum_ra_rX_{t-j-r}=\sum_\ell c_\ell X_{t-\ell},\qquad
\boxed{c_\ell=\sum_jb_ja_{\ell-j}.}
$$
Indeed $\sum_\ell|c_\ell|\leq\|a\|_1\|b\|_1$; this bound also justifies the interchange of the two $L^2$ sums. The <filter gain> multiplies under this composition. The sketches for the two individual filters are:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-47-filter-gains.png]
{title=Ordinary and twelve-step seasonal difference-filter gains}
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