Answer: no. Example: take a simple abelian surface X with real multiplication by Q($\sqrt{d}$) (where d is a square-free positive number). X has Picard number 2, and the intersection form on N^1(X) diagonalises to diag(1,-d). The nef cone is just the cone of classes x in N^1(X) with x^2 >= 0 (more precisely, the part of this cone which also satisfies x.h >=0 for a chosen ample class h), and a simple computation shows that the boundary rays of this cone are irrational (by square-freeness of d). Now for any abelian variety the effective cone is equal to the nef effective cone, which equals the union of the ample cone with the rational boundary faces of the nef cone. In our example the nef cone has no rational boundary faces, so the ample cone and the effective cone (minus zero) coincide.

Answer: no. Example: take a simple abelian surface X with real multiplication by Q($\sqrt{d}$) (where d is a square-free positive number). X has Picard number 2, and the intersection form on N^1(X) diagonalises over Q to diag(a,-b) where b/a=d. The nef cone is just the cone of classes x in N^1(X) with x^2 >= 0 (more precisely, the part of this cone which also satisfies x.h >=0 for a chosen ample class h), and a simple computation shows that the boundary rays of this cone are irrational (by square-freeness of d). Now for any abelian variety the effective cone is equal to the nef effective cone, which equals the union of the ample cone with the rational boundary faces of the nef cone. In our example the nef cone has no rational boundary faces, so the ample cone and the effective cone (minus zero) coincide.

Answer: no. Example: take a simple abelian surface X with real multiplication by Q($\sqrt{d}$) (where d is a square-free positive number). X has Picard number 2, and the intersection form on N^1(X) diagonalises to diag(1,-d). By simplicity the nef cone is just the cone of classes x in N^1(X) with x^2 >= 0 (more precisely, the part of this cone which also satisfies x.h >=0 for a chosen ample class h), and a simple computation shows that the boundary rays of this cone are irrational (by square-freeness of d). Now for any abelian variety the effective cone is equal to the nef effective cone, which equals the union of the ample cone with the rational boundary faces of the nef cone. In our example the nef cone has no rational boundary faces, so the ample cone and the effective cone (minus zero) coincide.

Answer: no. Example: take a simple abelian surface X with real multiplication by Q($\sqrt{d}$) (where d is a square-free positive number). X has Picard number 2, and the intersection form on N^1(X) diagonalises to diag(1,-d). The nef cone is just the cone of classes x in N^1(X) with x^2 >= 0 (more precisely, the part of this cone which also satisfies x.h >=0 for a chosen ample class h), and a simple computation shows that the boundary rays of this cone are irrational (by square-freeness of d). Now for any abelian variety the effective cone is equal to the nef effective cone, which equals the union of the ample cone with the rational boundary faces of the nef cone. In our example the nef cone has no rational boundary faces, so the ample cone and the effective cone (minus zero) coincide.