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.
Note that $\hat F$ is a convex function. It is also clear that $\hat F \ge 0$ since $F\ge 0$ and $\hat F(1) = 0$. First we have
$$\tag{1} \hat F(s) = 0, \ \ \ \ \text{ for all } s>1.$$
To show (1), assume that contrary that $\hat F(s_0) >0$ for some $s_0 >1$. Let $\ell(x) = ax+b$ be the linear function passing through $(1,0)$ and $(s_0, \hat F(s_0))$. Since $\hat F$ is convex, $F(s) \le \ell (s)$ for all $s\in (1, s_0)$. By convexity of $\hat F$ again, this implies
$$\tag{2} \hat F(s) \ge \ell(s), \ \ \ \ \text{for all } s> s_0.$$
Now since $F(s_0) >0$, the slope of $\ell$ is positive. Also
$$ \lim_{s\to +\infty} \frac{\sqrt 2 |\sqrt s-1|}{\ell(s)} = 0$$
Thus $\ell(s) >F(s)$ when $s$ is large enough. By (2) this is impossible since $\hat F \le F$. Thus we have shown (1).
It remains to find $\hat F(s)$ when $s <1$. Note that
\begin{align}
F'' \le 0 & \ \ \text{ on } [0,1/4), \\
F'' \ge 0 & \ \ \text{ on } (1/4,1).
\end{align}
and
$$\tag{3} \lim_{s\to 1/4^+} F'(s) = -\sqrt 2 < 2 (\sqrt 2 -2) = \frac{F(1/4) - F(0)}{1/4-0} $$
Let $y=\ell_1(s)$ be the straight line which passing through $(0,1)$ and is tangential to $F$ at some $s\in (1/4, 1)$. If $(\bar s, F(\bar s))$ is the point of contact, then
\begin{align}
\frac{F(\bar s) - 1}{\bar s-0} &= F'(\bar s)\\
\Rightarrow \sqrt 2 (1-\sqrt {\bar s})-1 &= -\bar s (\sqrt 2\sqrt {\bar s})^{-1} \\
\Rightarrow 2 - 2\sqrt {\bar s} - \sqrt 2&= -\sqrt {\bar s}\\
\Rightarrow \bar s &= (2-\sqrt 2)^2 = 6-4\sqrt 2.
\end{align}
Now we claim that
$$ h(s)= \begin{cases} \ell_1 (s) & \text{ if } s\in [0,\bar s], \\
F(s) & \text{ if } s\in (\bar s, 1],\\
0 & \text{ if } s >1.
\end{cases}.$$
Is the convex envelope. First of all, $h(s)$ is convex. Thus by defintion $h\le \hat F \le F$. If $h \neq \hat F$, then $h(s) < \hat F(s)$ for some $s\in (0, \bar s)$. But then $\hat F(s) > \ell_1(s)$, and $\ell_1$ is the line joining $(0,\hat F(0))$ and $\bar s, \hat F(\bar s))$, which contradicts the assumption that $\hat F$ is convex. Thus $h = \hat F$.
Best Answer
The only convex functions majorized by $\sin$ are the functions $h(x) = c$, where $c \le -1$. Hence $(\operatorname{conv} \sin)(x) = -1$.
To see this, suppose $h \le \sin$, $h$ is convex but $h$ is not constant. Then for some $x_1<x_2$, we have $h(x_1) \neq h(x_2)$. Suppose $h(x_1) < h(x_2)$, let $l(x) = \frac{x-x_2}{x_1-x_2}h(x_1) + \frac{x-x_1}{x_2-x_1}h(x_2)$, $l$ is affine and $\lim_{x \to \infty} l(x) = \infty$. Choose $x > x_2$, and write $x_2$ as a convex combination of $x_1,x$. This gives $x_2 = \frac{x_2-x_1}{x-x_1}x+\frac{x-x_2}{x-x_1}x_1$, and hence $h(x_2) \leq \frac{x_2-x_1}{x-x_1}h(x)+\frac{x-x_2}{x-x_1}h(x_1)$. Rearranging gives $l(x) \le h(x)$, which is a contradiction, as $h$ is bounded above by $1$ (since it is majorized by $\sin$). Hence $h$ is constant. Since $h \le \sin$, and $\sin (-\frac{\pi}{2}) = -1$, it follows that the constant is $\le -1$.
The following is a change from my previous answer, thanks to @byk7 (see the comments below) for catching a problem with my previous answer.
Here is another way of seeing this. Given a convex set $C \subset \mathbb{R}^n \times \mathbb{R}$, define $L_C(x) = \inf \{ \mu | (x,\mu) \in C \}$. It is straightforward to see (Theorem 5.3 Rockafellar) that $L_C$ is a convex function.
Now note that $\operatorname{conv} \sin=L_{\operatorname{co} ( \operatorname{epi} \sin)}$.
Since $\operatorname{co} ( \operatorname{epi} \sin) = \mathbb{R} \times [-1,\infty)$, we have $h(x) = \inf \{ \alpha | (\alpha,x) \in \operatorname{co} ( \operatorname{epi} \sin) \} = -1$.