is there some kind of characterization of $h$ such that the domain $Q$ is convex
No, because the support function of $Q$ is equal to the support function of the convex hull of $Q$.
is there some kind of characterization of $h$ such that the domain $Q$ is strictly convex
Yes: if and only if $h$ is differentiable. In other words, if and only if the polar set of $Q$ has smooth boundary. It's worth noticing that the polar of polar set is the original set.
A corner in a convex set creates a line segment in the boundary of its polar, because the value of support function comes from the same point (the corner) for some interval of $\theta$. Conversely, suppose the boundary of a convex set contains a vertical segment to the right of origin. Then $h$ is not differentiable at $\theta=0$, because a small change of $\theta$ in either direction increases $h$ at a linear rate. (So, the graph of $h$ has a nonsmooth minimum like $|\theta|$.)
The general form of the above is known as duality of smoothness and rotundity, and it holds in all finite dimensional spaces.
Is there any characterization of $h$ such that the boundary $\partial Q$ admits nonvanishing curvature?
Yes: if and only if $h\in C^2$. I recommend Convex Bodies: The Brunn-Minkowski Theory by Rolf Schneider, specifically section 2.5 Higher regularity and curvature. For the two-dimensional case, see also page 2 of Lectures on Mean Curvature Flows by Xi-Ping Zhu.
Sketch. Introduce the Gauss map $\nu:\partial Q\to S^1$ which gives exterior unit normal vector at every boundary point. This map is defined when $Q$ has $C^1$ boundary. If $\partial Q$ is also strictly convex, then $\nu$ is injective. And if $\partial Q$ has nonvanishing curvature, then $\nu$ is a diffeomorphism -- indeed, this is if and only if, because the derivative of $\nu$ is the curvature of $\partial Q$.
It is convenient to define $h_Q$ as a function on $\mathbb R^2$, via $h_Q(x)=\sup_{q\in Q} x\cdot q$. Then one can check that $ h_Q(x)= x\cdot \nu^{-1}(x/|x|) $ and
$ \nabla h_Q(x)= \nu^{-1}(x/|x|)$.
Thus, the equivalence is: nonvanishing curvature $\iff$ $\nu $ is a diffeomorphism $\iff$ $h_Q\in C^2(\mathbb R^2\setminus \{0\})$.
1. Yes, this is true. This can be derived from the involutive property of the Legendre-Fenchel transform: if $f:\mathbb R^n\to (-\infty,+\infty]$ is a closed convex function, then $f^{**}=f$ (Theorem 12.2 in Rockafellar's Convex Analysis). Recall that
$$f^*(x) = \sup\{\langle x,y\rangle - f(y) : y\in\mathbb R^n\}$$
and a convex function is closed if its epigraph is.
With every convex closed set $K$ we can associate a closed convex function $\delta_K$ by letting $\delta_K(x)=0$ when $x\in K$ and $\delta_K(x)=+\infty$ otherwise. Observe that $\delta_K^*$ is exactly $h_K$.
Given $h$ as in your question, define $K= \{x : \langle x,y\rangle \le h(y) \ \forall y\}$. Observe that $h^*(x)=0$ when $x\in K$, because the supremum is attained by $y=0$. If $x\notin K$, then there is $y$ such that $\langle x,y\rangle > h(y)$. Considering large multiples of such $y$, we conclude that $h^*(x)=\infty$.
Thus, $h^* = \delta_K$. By the involutive property, $h=h^{**} = \delta_K^* = h_K$.
By the way, this result is Theorem 13.2 in Rockafellar's book.
2. Yes, this is correct. Another way to state this fact: the epigraph of $h_{K_1\cap K_2}$ is the convex hull of $\operatorname{epi} h_{K_1} \cup \operatorname{epi} h_{K_2}$. To see this, observe that the epigraph of $h_K$ is determined by its intersection with the horizontal plane $z=1$. This intersection is nothing but $K^\circ$, the polar of $K$. It remains to use the fact that $(K_1\cap K_2)^\circ = \operatorname{conv}(K_1^\circ \cup K_2^\circ)$. See also Corollary 16.5.1 in Rockafellar's book.
Best Answer
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It might help to consider the indicator function. Also we will use the convention that column vectors are elements of $\R^d$ and row vectors are elements of the dual space $(\R^d)^*$ (even though $(\R^d)^*\cong \R^d$).
When $K\subseteq\R^d$ is a nonempty, closed and convex cone, then the support function $\sigma_K:(\R^d)^*\to\R\cup\{+\infty\}$ coincides with the conjugate function of the indicator of $K$ given by $\delta_K:\R^d\to\R\cup\{+\infty\}$ $$ \delta_K(x) = \begin{cases} 0,&\text{if $x\in K$} \\ +\infty,&\text{o.w.}\end{cases} = \delta_{K^\circ} $$ That is, $\sigma_K(\bullet) = \delta_K^*(\bullet) = \sup_x\{\bullet\cdot x - \delta_K(x)\} = \sup_{x\in K} \{\bullet \cdot x\}$. We can also see that $\sigma_K = \delta_{K^\circ}$ where $K^\circ = \{y:y\cdot x \leq 0,;\forall x\in K\}$ is the polar cone of $K$.
When $f$ is convex and smooth, then for any $y\in(\R^d)^*$, $f^*(y)$ can be interpreted as the signed distance between zero and the intercept of a supporting hyperplane of $f$ with slope $y$. We have this interpretation since in this special case, the slope of $f$ is monotone and hence there is a 1:1 map slopes $y$ and points $x$ (see here). However, when $f$ is not convex, we don't have this relationship, so the $\sup$ is taken to make this relationship well-defined.
We're in the latter case when $f=\delta_K$, which is nonconvex, and the only supporting hyperplane of $\delta_K$ is $y=0$ when $x\in K$ (if $x\notin K$, then there are no supporting hyperplanes). It helps to draw this out in 1d, for example with $K=\{x\in\R:x\geq0\}$ (see below).
This means that for any $y\in(\R^d)^*$, we can only interpret $\sigma_K(y)$ as the signed distance to zero of a supporting hyperplane for $\delta_K$ when $y=0$ and $x\in K$. Otherwise, there are no supporting hyperplanes, so the signed distance should be $+\infty$.