Analyse $f$. List a few values: $f(1)=1$, $f(2)=1$, $f(3)=2$, $f(4)=0$, $f(5)=3$, $f(6)=-1$, $f(7)=4$, $f(8)=-2$, and so on. Since $f(1)=f(2)$, $f$ is not one-one. As $n$ ranges over odd naturals, $f(n)=(n+1)/2$ covers all positive integers; as $n$ ranges over even naturals, $f(n)=(4-n)/2$ covers $1,0,-1,-2,\dots$, so all non-positive integers. Together, $f$ is onto $\mathbb{Z}$. (D) is correct.

Analyse $g$. For $n\ge 0$, $g(n)\in \{3,5,7,9,\dots \}$ (odd $\ge 3$). For $n<0$, $g(n)\in \{2,4,6,\dots \}$. The natural number $1$ is never attained, so $g$ is not onto. It is one-one (each branch is injective and the images are disjoint), but not onto. (C) is false.

Analyse $f\circ g$. For $n\ge 0$, $g(n)=2n+3$ is odd, so $f(g(n))=(2n+4)/2=n+2$. For $n<0$, $g(n)=-2n$ is a positive even integer, so $f(g(n))=(4-(-2n))/2=2+n$. Hence $f\circ g(n)=n+2$ on all of $\mathbb{Z}$, which is a bijection. So $f\circ g$ is one-one. (B) is false.

Analyse $g\circ f$. $g(f(1))=g(1)=5$ and $g(f(2))=g(1)=5$, so $g\circ f$ is not one-one. Range: $g$ outputs only naturals that are $\ge 2$ from one branch and $\ge 3$ odd from the other, so $1$ is never attained; hence $g\circ f$ is not onto $\mathbb{N}$. (A) is correct.

$\therefore$ The correct options are (A) and (D).