Q1: In general no: take $X$ to be the subvariety of $P^{\delta_d\vee}$, the dual of $P^{\delta_d}$, formed by all $H_p$'s. Note that $X$ is just the image of the Veronese map $\mathbb{P}(V)\to\mathbb{P}(Sym^d(V^\vee))$ for $V=\mathbb{C}^3$. So $X$ lies on a quadric $Q$ given by $x_i x_j=x_kx_l$ with $i,j,k,l$ pairwise disjoint. Here $x_i,x_j,x_k,x_l$ are coordinate functions on $(Sym^d \mathbb{C}^3)^\vee$. Set $\mathcal{A}=Q^\vee$, the dual of $Q$; this is a 2-dimensional quadratic surface in $\mathcal{D}$. By the biduality theorem, $\mathcal{A}^\vee=Q$. So $X\subset Q$ is contained in the dual of $\mathcal{A}$, which means that for every $H_p\in X$ the intersection $H_p\cap \mathcal{A}$ (and hence also $H^*_p\cap\mathcal{A}$) is not transversal.
Q2: Yes, for any $p$, since both $\mathcal{A}$ and $\partial H_p$ are (quasiprojective) algebraic varieties.
Q3: In general no: take $\mathcal{A}$ to be the smooth part of the discriminant hypersurface $\Delta\subset P^{\delta_d}$ formed by projectivizing the set of all homogeneous polynomials $f$ such that the gradient of $f$ vanishes at some point of $\mathbb{P}^2$. Then for any $p$ we have $\dim \partial H_p\cap \mathcal{A}=\dim\partial H_p=\delta_d-3\neq k-3=\delta_d-4$.
Re questions 3 and 4: let me describe informally what one should expect. One of the two things can happen. Either $\mathcal{A}\subset\Delta$, in which case $\dim \mathcal{A}\cap\Delta=\dim\mathcal{A}=k$ and for generic $p$ one would expect $\dim \partial H_p\cap \mathcal {A}$ to be $k-2$ since $\partial H_p$'s form a 2-parametric family of projective subspaces. Or $\mathcal{A}\not\subset\Delta$, in which case $\dim \mathcal{A}\cap\Delta=\dim\mathcal{A}=k-1$ and for generic $p$ one would expect $\dim \partial H_p\cap \mathcal {A}$ to be $k-3$