= Solution
Use the printed parameterization $\lambda=\sigma^2$, $v=\sigma_0^2$, and $a=\sqrt\rho$. Let $T=365$, $X$ be the <design matrix> with rows $x_i^T$, and $r=\theta-X\beta$. Conditional on $\theta$ and the parameters, $Y$ contributes no extra dependence on $Z$ by <conditional independence>.
The stationary autoregressive <normal distribution> has prior density proportional to
$$
\exp\left[-\frac12\left(z_1^2+\frac1{1-a^2}\sum_{i=2}^T(z_i-az_{i-1})^2\right)\right].
$$
Its <Gaussian autoregressive conditional precision> is the <tridiagonal matrix> $J$ with
$$
J_{11}=J_{TT}=\frac1{1-a^2},\quad
J_{ii}=\frac{1+a^2}{1-a^2}\ (1<i<T),\quad
J_{i,i+1}=J_{i+1,i}=-\frac a{1-a^2}.
$$
Combining this prior with $r\mid Z\sim N(\lambda Z,vI)$ and completing the square gives
$$
\boxed{Z\mid\beta,\sigma^2,\sigma_0^2,\rho,\theta,x,Y
\sim N(Q^{-1}h,Q^{-1}),\quad Q=J+\frac{\sigma^4}{\sigma_0^2}I,\quad h=\frac{\sigma^2}{\sigma_0^2}(\theta-X\beta).}
$$
The <precision matrix> $Q$ is positive definite. Compute its lower <Cholesky decomposition> $Q=LL^T$, solve $Qm=h$, draw $\xi\sim N(0,I_T)$, and solve $L^T\eta=\xi$. Then return $Z=m+\eta$. This <Gaussian sampling from a precision Cholesky factor> is exact because $\operatorname{Cov}(\eta)=L^{-T}L^{-1}=Q^{-1}$. Tridiagonal structure gives linear cost in $T$ and preserves the conditional correlation between days. Notice that the solve uses $L^T$, not $L$.
These conditionals are normalized for every fixed positive $\sigma^2,\sigma_0^2$, despite the joint posterior impropriety explained in part (a). If the intended model uses $\sigma Z_i$ instead, replace the observation coefficient $\lambda$ by $\sigma$ throughout; the information and precision terms then use $\sigma/v$ and $\sigma^2/v$.
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