If $f$ is a function $A\to A$ where $A$ is a finite set, then $f$ is injective if and only it it is surjective. It is not possible for such an $f$ to be injective without being surjective or vice versa.
This can be proved by induction on the number of elements in $A$, but the details are slightly tedious. This property carries over to a function $f:A\to B$ where $A$ and $B$ have the same number of elements -- because "the same number of elements" means that there is a bijection $g:A\to B$, and then $f$ is injective/surjective exactly when $g^{-1}\circ f$ is, and $g^{-1}\circ f$ goes from $A$ to $A$ itself.
On the other hand, if $A$ is infinite, then it is possible for a function to be injective without being surjective, or vice versa. Some simple examples for $\mathbb N\to\mathbb N$ is
$$ g(n) = n+7 \\
h(n) = \begin{cases} n & \text{if }n\le 42 \\ n-1 & \text{otherwise} \end{cases}$$
where $g$ is injective but not surjective, and $h$ is surjective but not injective.
(Strictly speaking this property of infinite sets is only guaranteed if the axiom of choice holds, but that is a techincal complication that you probably don't want to worry about at this level).
The composite function $f \circ g$ is only defined if the codomain of $g$ is contained in the domain of $f$. Depending on the definition you are using, you may need the stronger condition that the codomain of $g$ equals the domain of $f$.
In this case, the codomain of $g$ is $\mathbb{N}$ and the domain of $A = \{0, 1\}$. Moreover, as noted in the comments, since the range of $g$ is $\{6, 8, 10, 12\}$, the elements in the range of $g$ are not contained in the domain of $f$. Thus, the composite function $f \circ g$ is undefined.
Since $f: A \to B$ and $g: B \to C$, the composite function $g \circ f$ is defined since the codomain of $f$ equals the domain of $g$. As you correctly concluded, the function $g \circ f: A \to C$ is defined by
$$(g \circ f)(x) = g(f(x)) = g(x + 5) = 2(x + 5) - 2 = 2x + 8$$
The graph of the composite function $g \circ f$ is the set of all ordered pairs $(a, c) \in A \times C$ such that $(g \circ f)(a) = c$ for some $a \in A$. Since $(g \circ f)(0) = 8$ and $(g \circ f)(1) = 10$, the graph of the function $g \circ f$ is the set
$$G_{g \circ f} = \{(0, 8), (1, 10)\}$$
Since $g: B \to C$ and $h: C \to A$, the composite function $h \circ g$ is defined since the codomain of $g$ equals the domain of $h$. Since $h: C \to A$ is defined by
$$h(x) = \begin{cases}
0 & \text{if $x < 10$}\\
1 & \text{if $x \geq 10$}
\end{cases}
$$
and $g: B \to C$ is defined by $g(x) = 2x - 2$, observe that if $x \in B$, then
$$(g \circ h)(x) = \begin{cases}
0 & \text{if $x < 6$}\\
1 & \text{if $x \geq 6$}
\end{cases}
$$
To see this, observe that if $x < 6$, $g(x) = 2x - 2 < 2 \cdot 6 - 2 = 10$, and that if $x \geq 6$, $g(x) = 2x - 2 \geq 2 \cdot 6 - 2 = 10$. Since
\begin{align*}
(h \circ g)(4) & = h(g(4)) = h(6) = 0\\
(h \circ g)(5) & = h(g(5)) = h(8) = 0\\
(h \circ g)(6) & = h(g(6)) = h(10) = 1\\
(h \circ g)(7) & = h(g(7)) = h(12) = 1
\end{align*}
the graph of $h \circ g$ is
$$G_{h \circ g} = \{(4, 0), (5, 0), (6, 1), (7, 1)\}$$
Observe that $h: C \to A$ and $g: B \to C$, so the codomain of $h$ is not contained in the domain of $g$ since $A$ is not a subset of $B$. Thus, the composite function $g \circ h$ is not defined.
Best Answer
The range is characterized as the set of all real numbers $y$ such that $f(x)=y$ for at least one $x$ in the domain of $f$.
[The domain consists of all $x$ with $16-x^{2} \geq 0$ which means $-4 \leq x \leq 4$. The range consists of all non-negative real numbers less than or equal to $4$. [If $0 \leq y \leq 4$ then $x=\sqrt {16-y^{2}}$ satisfies $f(x)=y$].