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EDIT Added a concrete example for the answer for Q4.

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 greater greate\r difference is possible. Take a general K3 surface: $NE(X)$ is $1$-dimensional and $H^2$ is $22$.

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.

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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 greater 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.

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)).

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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. Then

The long exact cohomology sequence of the answers are:

1) YES, because now exponential sequence $$ 0\to \mathbb Z \to \mathscr O_X \to \mathscr O_X^* \to 0 $$ Shows that the kernel of that the map ${\rm Pic} X\to H^2(X,\mathbb Z)$ is actually ${\rm Pic}^\circ X$which . By the above this means that ${\rm Pic} X\to H^2(X,\mathbb Z)$ is trivialinjective 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 NOno. Mostly NOno. 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 greater difference is possible. Take a general K3 surface: $NE(X)$ is $1$-dimensional and $H^2$ is $22$.

3) YESYes, by the first paragraphtwo paragraphs.

4) NONo. Or YESyes, 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.
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