= Solution
A random <vector> $X$ has <multivariate normal distribution> $N_d(\mu,\Sigma)$ if every real linear combination $t^TX$ is normally distributed, with mean $t^T\mu$ and variance $t^T\Sigma t$. The <covariance matrix> is symmetric positive semidefinite; zero variances permit degenerate normals. Its <moment-generating function> is $M_X(t)=\exp(t^T\mu+t^T\Sigma t/2)$, including the degenerate cases.
For the given $X$, let $t_j\in\mathbb R^{d_j}$. The joint <moment-generating function> of the $Y_j=A_jX$ is
$$
\begin{aligned}
\mathbb E\exp\left(\sum_jt_j^TY_j\right)
&=M_X\left(\sum_jA_j^Tt_j\right)\\
&=\exp\left[\frac{\sigma^2}{2}\sum_{i,j}t_j^TA_jA_i^Tt_i\right]\\
&=\prod_j\exp\left[\frac{\sigma^2}{2}t_j^TA_jA_j^Tt_j\right].
\end{aligned}
$$
The last equality uses the assumed orthogonality of the row spaces. Its factors are exactly the individual <moment-generating functions>, so the hint implies \b[the random <vectors> $Y_j$ are independent]. Also every linear combination of the stacked <vector> $Y$ is a linear combination of $X$, hence normally distributed. Consequently
$$
\boxed{Y\sim N_{\sum_jd_j}\left(0,\operatorname{diag}(\sigma^2A_1A_1^T,\ldots,\sigma^2A_JA_J^T)\right).}
$$
This establishes both claims without excluding singular block covariances.
For the <sample mean> and <sample variance> result, assume $n\ge2$ and set $W=(Z_1-\mu,\ldots,Z_n-\mu)^T$. Independence of the observations gives $W\sim N_n(0,\sigma^2I)$. Let $u=(1,\ldots,1)^T$, $A_1=u^T/n$, and $A_2=I-uu^T/n$. Then $A_1A_2^T=0$. The first transformed <vector> is the scalar $\overline Z-\mu$, and the second is the residual <vector> $(Z_i-\overline Z)_i$. They are independent by the just-proved result. Applying the measurable function $v\mapsto\|v\|^2/(n-1)$ to the residual <vector> preserves independence from the <sample mean>. Thus
$$
\boxed{\overline Z\ \text{and}\ S_{ZZ}\ \text{are independent}.}
$$
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