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.
Your proof is incorrect because $\Phi$ might not be extendable to a convex function on entire $\Bbb{R^n}$. For example take $\Phi$ the function whose graph is the lower half of the unit circle.
Now how to proof what you claimed: Subdifferentials and directional derivatives is a local feature of function. So first prove that $ y \in \partial \Phi (p) $ if and only if there exist an open neighborhood of $p$, say $X$ such that
$$\ \langle y, q-p \rangle \leq \Phi(q) - \Phi(p) \quad \ \forall q \in X $$
Hint: For right to left define the function $ f(x)= \Phi(x) - \Phi(p) - \langle y, x-p \rangle$ observe that $f$ is convex on the whole $\Bbb{R^n}$ and take a local minimum at $x =p$, so it has to be global minimum too.
Now mimic the Bertsekas' proof, for the local version of subdifferential .
Best Answer
The first is true. The inclusion $\overline{\operatorname{int} K_1} \subset K_1$ holds for all closed $K_1$, convex or not. To see the other inclusion, fix $a \in \operatorname{int} K_1$. For every $b \in K_1$, $$\gamma_b \colon t \mapsto a + t(b-a),\quad t \in [0,1]$$ is a path in $K_1$ connecting $a$ and $b$, and $\gamma_b(t) \in \operatorname{int} K_1$ for $t \in [0,1)$ since $$(1-t)\cdot (\operatorname{int} K_1) + tb$$ is an open neighbourhood of $\gamma_b(t)$ contained in $K_1$. Then $b = \lim_{t \to 1} \gamma_b(t)$ shows $b \in \overline{\operatorname{int} K_1}$.
The second doesn't hold. Trivial counterxample: $K_1 = \varnothing$ and $K_2 = X$.
Slightly less trivial counterexample: Let $H$ be a closed hyperplane(1) in $X$ (if that doesn't exist, the trivial counterexample might be the only one in $X$). We can take $K_1$ to be one of the closed half-spaces determined by $H$ and $K_2$ to be either $H$, or the other closed half-space determined by $H$. Then $\partial K_1 = H = \partial K_2$, but the exteriors are different ($\operatorname{int} K_1$ is contained in $\operatorname{ext} K_2$ and not empty).
I don't know if there are other counterexamples. I couldn't think of any, but neither do I see a proof that there are no others.
(1) Viewing $X$ as a real vector space, even if it also is a complex vector space.