Suppose $Tx=y$, we have:
$$y(t)=(Tx)(t)= \int_0^t x(r) dr$$
Since $x$ is continuous, $y$ is differentiable and we have:
$$x=y'$$
So
$$T^{-1}(y)=y'$$
About range of $T$, it's clear that each $Tx$ is differentiable and
$$(Tx)(0)=0$$
Let $y\in C[0,1]$ be a differentiable function such that
$$y(0)=0$$
we have:
$$(Ty')(t)=\int_0^t y'(r) dr=y(t)$$
provided this integral exists. So which integral do you use? e.g if this definition is used then the integral always exists and the range of $T$ is all differentiable functions that are 0 at 0.
About Riemann integration I think the range is more limited.
Since every $x \in C[a, b]$ is continuous on a compact set, it's bounded.
For any $x \in C[a, b]$, we have:
$$
|f(x)| = \left|\int_a^b x(t) x_0(t) \, dt \right| \le \int_a^b |x(t)| \cdot |x_0(t)| \, dt \le \|x\| \int_a^b |x_0(t)| \, dt
$$
Thus, $f$ is bounded and its norm satisfies:
$$
\|f\| \le \int_a^b |x_0(t)| \, dt \newcommand{sgn}{\operatorname{sgn}}
$$
In fact, equality holds. To see this, consider the sign function $\hat x(t) = \sgn(x_0(t))$. By Lusin's theorem, there is exists a sequence of functions $x_n \in C[a, b]$ such that $\|x_n\| \le 1$ and $x_n(t) \to \hat x(t)$ as $n \to \infty$ for every $t \in [a, b]$. By the dominated convergence theorem, we have:
\begin{align}
\lim_{n \to \infty} f(x_n) &= \lim_{n \to \infty} \int_a^b x_n(t) x_0(t) \, dt \\
&= \int_a^b \lim_{n \to \infty} x_n(t) x_0(t) \, dt \\
&= \int_a^b \sgn(x_0(t)) x_0(t) \, dt \\
&= \int_a^b |x_0(t)| \, dt
\end{align}
For the second linear functional you added, approximate the following function via Lusin's theorem in a similar manner:
$$
\hat x(t) = \begin{cases} 1 & \text{if } t \le \frac{a+b}{2} \\
-1 & \text{if } t > \frac{a+b}{2}
\end{cases}
$$
Best Answer
It is obvious that $\Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty$, hence $\Vert T\Vert\leq 1$. Since $\Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty$, we conclude that $\Vert T\Vert=1$.
One can easily verify that $$T(l^\infty)=\left\{(\xi_j):\exists C>0 \text{ such that } |\xi_j|\leq C/j\text{ for all j}\right\}$$
$T(l^\infty)$ is not closed in $l^\infty$: Let $x_n=(1,\dfrac{1}{\sqrt{2}},\ldots,\dfrac{1}{\sqrt{n}},0,0,\ldots)=T(1,\sqrt{2},\ldots,\sqrt{n},0,\ldots)$. The sequence $(x_n)$ obviously converges to $x=\left(1/\sqrt{j}\right)_{j=1}^\infty\in l^\infty$, which does not lie in $T(l^\infty)$ (if it did, what would be it's pre-image?).
The inverse operator $T^{-1}$ is not bounded: Consider the sequence $(x_n)\subseteq T(l^\infty)$ as above. This sequence is bounded but the image $\left\{T^{-1}x_n\right\}$ is not, since $\Vert T^{-1}x_n\Vert_\infty=\sqrt{n}$.
You could also argue in this way: If $T^{-1}$ were bounded, then for every Cauchy sequence $(y_n)\in T(l^\infty)$, the sequence $(T^{-1} y_n)$ would also be Cauchy, and hence would converge to some $x\in l^\infty$ since $l^\infty$ is complete. But then, since $T$ is bounded, $y_n$ would converge to $Tx$. Then $T(l^\infty)$ would be complete, contradicting the fact that it is not closed in $l^\infty$ (this argument can actually be used to show that if $T:X\rightarrow Y$ is an isomorphism between a Banach space $X$ and some normed space $Y$ such that $T$ and $T^{-1}$ are bounded, then $Y$ is Banach).