If $f(x)=\{\begin{array}{l}2+2x,-<!--\; -\; -->1\le <!--\; \le \; -->x<0\\ 1-<!--\; -\; -->\frac{x}{3},0\le <!--\; \le \; -->x\le <!--\; \le \; -->3\end{array}$; $g(x)=\{\begin{array}{l}-<!--\; -\; -->x,-<!--\; -\; -->3\le <!--\; \le \; -->x\le <!--\; \le \; -->0\\ x,0<x\le <!--\; \le \; -->1\end{array}$, then range of $(fog(x))$ is

The correct answer is [0, 1].

Let us determine the range of the composite function $(f\circ g)(x)$ step-by-step.

First, let's analyze the inner function $g(x)$, which is defined as:
$g(x)=\{\begin{array}{l}-x,-3\le x\le 0\\ x,0<x\le 1\end{array}$

Let's find the range of $g(x)$ over its domain:
1. For $-3\le x\le 0$, we have $g(x)=-x$. Multiplying the inequality by -1 gives $0\le -x\le 3$, so the range here is $[0,3]$.
2. For $0<x\le 1$, we have $g(x)=x$, so the range here is $(0,1]$.

Combining the two intervals, the overall range of $g(x)$, which serves as the input domain for the outer function $f$ in the composition $f(g(x))$, is:
$\text{Range of}g=[0,3]$

Now we evaluate $f(t)$ for $t\in [0,3]$, where $t=g(x)$. Let us look at the definition of $f(t)$:
$f(t)=\{\begin{array}{l}2+2t,-1\le t<0\\ 1-\frac{t}{3},0\le t\le 3\end{array}$

Since the input value $t$ lies in the interval $[0,3]$, it falls completely under the second branch of $f(t)$:
$f(t)=1-\frac{t}{3}$

Let us compute the range of $1-\frac{t}{3}$ for $0\le t\le 3$:
1. Start with the domain inequality:
$0\le t\le 3$
2. Divide all terms by 3:
$0\le \frac{t}{3}\le 1$
3. Multiply by -1 (reversing the inequality signs):
$-1\le -\frac{t}{3}\le 0$
4. Add 1 to all parts of the inequality:
$1-1\le 1-\frac{t}{3}\le 1+0$
Which simplifies to:
$0\le f(t)\le 1$

Thus, the range of $(f\circ g)(x)$ is $[0,1]$.