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The answer to $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed, assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$, where $Z$ is effective and numerically trivial on $X$; so $Z=0$ and $C=D$. In particular, no positive linear combination of smooth curves can be numerically equivalent to $C$.

This also shows that the answer to $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.

The answer to question $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed, assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$, where $Z$ is effective and numerically trivial on $X$; so $Z=0$ and $C=D$. In particular, no positive linear combination of smooth curves can be numerically equivalent to $C$.

This also shows that the answer to question $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.

The answer to $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed, assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$ where $Z$ is numerically trivial on $X$, but $X$ is a blow-up of $\mathbb{P}^2$ so $Z=0$ and $C=D$. In particular, no positive linear combination of smooth curves can be numerically equivalent to $C$.

This also shows that the answer to $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.

The answer to $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed, assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$, where $Z$ is effective and numerically trivial on $X$; so $Z=0$ and $C=D$. In particular, no positive linear combination of smooth curves can be numerically equivalent to $C$.

This also shows that the answer to $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.

The answer to $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$ where $Z$ is numerically trivial on $X$, but $X$ is a blow-up of $\mathbb{P}^2$ so $Z=0$ and $C=D$.

This also shows that the answer to $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.

The answer to $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed, assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$ where $Z$ is numerically trivial on $X$, but $X$ is a blow-up of $\mathbb{P}^2$ so $Z=0$ and $C=D$. In particular, no positive linear combination of smooth curves can be numerically equivalent to $C$.

This also shows that the answer to $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.