The key here is the simple change-of-interval/rescaling formula, found e.g. at the link in the OP, according to which
\begin{equation}
T_n(f)(x)=T_{n,[0,x]}(f)=x\sum_1^n w_i f(xx_i), \tag{*}
\end{equation}
where the $w_i$'s and $x_i$'s are such that $T_n(f)(1)=T_{n,[0,1]}(f)=\sum_1^n w_i f(x_i)$ for all $f$.

From here, one immediately has the affirmative answer to Questions (1), (2), and (4). In particular, in (4), one may take $M_n=\sum_1^n|w_i|=\sum_1^n w_i =1$.

The answer to Question (5) is also positive. Indeed, recall that for any polynomial $p$ of degree $\le2n-1$ we have $T_n(p)(1)=\int_0^1 p$ and hence, by the rescaling, $T_n(p)(x)=\int_0^x p$ for all $x\in[0,1]$. Take now any $f\in C[0,1]$ and any real $\delta>0$ and let $p$ be a polynomial, of degree $\le2n-1$ for some natural $n$, such that $p-\delta\le f\le p+\delta$ on $[0,1]$. The operator $T_n$ is positive, since $w_i\ge0$ for all $i$. So, for $x\in[0,1]$
\begin{multline}
T(p)(x)-\delta\cdot x=\int_0^x (p-\delta)=T_n(p-\delta)(x)\le T_n(f)(x) \\
\le T_n(p+\delta)(x)=\int_0^x (p+\delta)
=T(p)(x)+\delta\cdot x,
\end{multline}
whence
\begin{equation}
|T_n(f)(x)-T(f)(x)|\le\delta\cdot x\le\delta,
\end{equation}
so that $T_n(f)\to T(f)$ uniformly on $[0,1]$, for any $f\in C[0,1]$.

Moreover, if two functions, say $f$ and $g$ in $C[0,1]$, are uniformly close to each other, than any polynomial $p$ uniformly close to $f$ will also be uniformly close to $g$. Hence, it follows from the above reasoning that $T_n\to T$ in the $(\|\cdot\|_\infty,\|\cdot\|_\infty)$ operator norm.

The answer to Question (3) is this: the linear operator $T_n$ is not injective for any natural $n$. Indeed, letting, as usual, the nodes $x_i$ be increasing in $i$, we have $x_n<1$. On the other hand, by (*), $T_n(f)=0$ for any $f\in C[0,1]$ such that $f=0$ on $[0,x_n]$. So, letting e.g. $f(x):=\max(0,x-x_n)$ for $x\in[0,1]$, we have $f\in C[0,1]$ and $f\ne0$, whereas $T_n(f)=T_n(0)$.

**Added in response to a comment by the OP:** No, the operator $T_n$ is not compact for any natural $n$. Indeed, take any compact intervals $J$ and $K$ such that $\emptyset\ne J^\circ\subset J\subset K^\circ\subset K\subset (x_{n-1}/x_n,1)$, where ${}^\circ$ denotes the interior and $x_{n-1}:=0$ for $n=1$. Take any $h\in C(J)$. Then $h$ equals $g|_J$, the restriction to $J$ of some function $g\in C[0,1/x_n]$ such that the support $\text{supp}\,g$ of $g$, defined as the closure of the set $\{x\colon g(x)\ne0\}$, is contained in $K$ and $\|g\|_{[0,1/x_n]}=\|h\|_J$, where $\|u\|_I:=\max_{x\in I}|u(x)|$ for any interval $I$ and any $u\in C(I)$.
Let now
\begin{equation}
f(y):=\frac{x_n}{w_n y}\,g\Big(\frac y{x_n}\Big)
\end{equation}
for $y\in(0,1]$, with $f(0):=0$. Then $\text{supp}f\subseteq x_n\,K\subset(x_{n-1},x_n)\subset(0,1)$, whence $y_*:=\min\text{supp}f\ge\min(x_n K)=x_n\min K>0$, $f\in C[0,1]$, and
\begin{equation}
\|f\|_{[0,1]}\le M\|g\|_{[0,1/x_n]}=M\|h\|_J,
\end{equation}
where $M:=\frac1{w_ny_*}<\infty$.
Next, for $x\in(0,1]$,
\begin{equation}
xw_i f(xx_i)=xw_i \frac{x_n}{w_n xx_i}\,g\Big(\frac{xx_i}{x_n}\Big)=
\left\{
\begin{aligned}
0&\text{ if }i\le n-1,\\
g(x)&\text{ if }i=n,
\end{aligned}
\right.
\end{equation}
because $\text{supp}\,g\subseteq K\subset (x_{n-1}/x_n,1)$ and hence $g=0$ on $[0,x_{n-1}/x_n]$.
So, $T_n(f)=g$ on $[0,1]$ and hence $T_n(f)=h$ on $J$, that is, $(R_JT_n)(f)=h$, where $R_J\psi:=\psi|_J$, the restriction of $\psi$ to $J$. So, $(R_JT_n)(MB_{[0,1]})\supseteq B_J$, where $B_I$ denotes the unit ball in $C(I)$. Since $B_J$ is not compact, it follows that the operator $R_JT_n$ is not compact. Therefore and because the operator $R_J$ is continuous, it follows that the operator $T_n$ is indeed not compact.