Here's a hint for (1): think about scaling the first $n$ rows by $-i$.
Still thinking about (2). I haven't worked this out, but I think considering the signed permutations way of calculating the determinant, and pairing $\sigma$ with $\sigma^{-1}$ might be helpful.
edit: Yeah, okay, just got back, have something kinda similar to levap. A matrix of the form $D$ is exactly a complex linear one, under the identification $\mathbb{R}^{2n} \rightarrow \mathbb{C}^n$, where $\mathbb{R}^{2n}$ has ordered basis $x_1, \ldots, x_n, y_1, \ldots, y_n$ and $\mathbb{C}^n$ $z_i = x_i + i y_i$.
With this, one knows as an operator on $\mathbb{C}^n$ one can write it in Jordan canonical form. If one has a Jordan block: $\begin{pmatrix} \lambda & - & \ldots \\ & \lambda & \ldots \end{pmatrix}$with basis $w_i$, setting $w_i = x'_i + iy'_i$, in the ordered basis $x_1', y_1', x_2', y_2' \ldots$, it's easy to check that $D$ has determinant $|\lambda|^r$, where $r$ is the size of the Jordan block.