I don't think that your assertion is true; for example, Lazarsfeld gives an example (PAG, 2.3.3) of a big and nef divisor on a surface such that its graded algebra is not finitely generated, so that the divisor can't be semiample.
But there are some close results for nef and big divisors, or even for good divisors (when the Kodaira dimensions equals the numerical dimension) as Mourougane and Russo showed : for example, Wilson's theorem asserts that for any nef and big divisor on an irreducible projective variety, there exists $m_0\in \mathbb N$ together with an effective divisor $N$ such that for all $m\geq m_0$, the linear system $|mD-N|$ has no base-point. (PAG, 2.3.9)
EDIT Added a concrete example for the answer for Q4.
By a result of Kleiman (somewhere in SGA6 and also in Lazarsfeld's book around 1.1.20) a numerically trivial line bundle has a power that's in ${\rm Pic}^\circ X$ which is trivial if $X$ is simply connected, so on such an $X$ a numerically trivial line bundle is necessarily torsion, but that corresponds to a finite étale cover, which is again trivial by the simple connectedness assumption, so there are no numerically trivial line bundles.
The long exact cohomology sequence of the exponential sequence
$$
0\to \mathbb Z \to \mathscr O_X \to \mathscr O_X^* \to 0
$$
Shows that the kernel of the map ${\rm Pic} X\to H^2(X,\mathbb Z)$ is ${\rm Pic}^\circ X$. By the above this means that ${\rm Pic} X\to H^2(X,\mathbb Z)$ is injective and since there are no torsion line bundles, it remains injective after tensoring with $\mathbb R$.
Then the answers are:
1) Yes, if $X$ is a surface since curves are divisors. In general, without the simple connectedness assumption the answer is certainly no, but that was not what you asked.
2) Actually no. Mostly no. Many examples, perhaps the simplest one is any (smooth projective) curve with $b_2>1$ or any smooth projective surface: $NE(X)$ will land in $H^{1,1}$ but $H^{2,0}\neq 0$. But even greate\r difference is possible. Take a general K3 surface: $NE(X)$ is $1$-dimensional and $H^2$ is $22$.
3) Yes, by the first two paragraphs.
4) No. Or yes, it is possible to have a line bundle like that. Let $\pi: X\to Z$ be a flat morphism from a smooth projective $4$-fold $X$ to a smooth projective $3$-fold $Z$, for instance a $\mathbb P^1$-bundle. Let $H\subset X$ be the pull-back of an ample line bundle from $Z$. Now if $Y$ is a general hyperplane section of $X$ that does not contain any fibre of $\pi$, then the induced map $Y\to Z$ is finite and hence $H\cap Y$ is an ample divisor.
Here is a concrete example: Let $Z$ be an arbitrary smooth projective threefold and let $\mathscr E$ be a rank $2$ vector bundle on $Z$ such that there exists a surjective map $\mathscr E\to \mathscr L$ onto a line bundle on $Z$. Consider $\pi: X=\mathbb P(\mathscr E)\to Z$ and observe that after twisting by the pull-back of a sufficiently ample line bundle on $Z$ we may assume that $\mathscr O_{\mathbb P(\mathscr E)}(1)$ is very ample. The surjective morphism $\mathscr E\to \mathscr L$ produces a section of $\pi$ and let us denote the image of that with $Y$. Just for kicks, notice that $Y\simeq Z$ so it can be any smooth projective variety and $Y\cap H$ corresponds to the divisor that pulls back to $H$ so it is indeed ample on $Y$. Also notice that $Y$ is a divisor on $X$ representing $\mathscr O_{\mathbb P(\mathscr E)}(1)$, so by the above assumption it is a hyperplane section of $X$. Choosing $\mathscr E\to \mathscr L$ generally makes $Y$ general but this is actually not necessary for the construction. Since $Y$ is a section, it intersects every fiber of $\pi$ in exactly $1$ point, so it cannot contain a fibre.
3.5) (inspired by Dave's answer): An interesting related problem is the connection between algebraic and homological equivalence. Again algebraic implies homological equivalence, but by a result of Griffiths (On the periods of certain rational integrals. I, II. Ann. of Math. (2) 90 (1969), 460-495; (2) 90 1969 496–541) they are not the same and in fact the difference can be quite large as proved by Clemens (Homological equivalence, modulo algebraic equivalence, is not finitely generated.
Inst. Hautes Études Sci. Publ. Math. No. 58 (1983), 19–38 (1984)).
Best Answer
A more direct approach is the following:
Let $X$ be the projective bundle $\pi:\mathbb{P}(\mathcal{E})\to \mathbb{P}^1$ where $\mathcal{E}=\mathcal{O}\oplus \mathcal{O}(-1) \oplus \mathcal{O}(-2)$. Let $M$ be the tautological bundle of $X$. It is easily checked that the ample line divisors $H_i$ on $X$ correspond to line bundles of the form $M^a\otimes \pi^*\mathcal{O}(b)$ with $b>2a$. We show that $Mov(X)$ is not spanned by products of the form $H_1\cdot H_2$.
Consider the line bundle $L=M\otimes \pi^*\mathcal{O}(-1)$. Using the Leray spectral sequence for the morphism $\pi$ we easily see that $L$ is not pseudoeffective. However, it is also straightforward to check that $$L\cdot H_1\cdot H_2=b_1+b_2-4>0$$ for $H_i=M\otimes \pi^*\mathcal{O}(b_i)$. Hence $L$ lies in the dual cone of $\overline{Q}(X)$ (using your notation). Now, if $\overline{Mov}(X)$ was generated by the $H_1\cdot H_2$'s this would imply that $L$ is pseudoeffective (by BDPP), a contradiction.