So we are seeking interpretation in terms of the Ito's integral, whose definition, as we know from the comments below, is in the sense of adapted process. This is not the end though, one can extend the Ito for non-adapted processes- e.g., Skorokhod which replaces the adaptability by regularity condition, and one can understand this integral intuitively in terms of Riemann sum and step processes. In essence one can extend Ito's integral to non-adaptive processes, the processes have to satisfy some conditions, but not going to go there!
The answer to one could vary depending on the interpretation one uses. Here is one way to go about it:
$\int_0^TW_TdW_t=\int_0^T\int_0^TdW_s\,dW_t$
$=2\int_0^T\int_0^tdW_s\,dW_t-\int_0^T{dW_s^2}$
$={2\int_0^T\int_0^t{dW_s\,dW_t}}-T$
I think it should equal $2\int_0^TW_t\,dW_t+T$ in the Ito's sense. On the other hand, if one tries a slightly different interpretation when approximating the integral via finite sum (think $n \to \infty$ in the partition sense etc.)
$\int_0^TW_TdW_t=\int_0^T\left(W_T-W_t\right)dW_t+\int_0^T W_tdW_t$
$={ \sum_{k=1}^{n}{\left( W_{t_{n}} - W_{t_{k}} \right) \Delta W_{t_{k}} }}+\int_0^T W_tdW_t$
$={ \sum_{k=1}^{n}{\left( W_{t_{n}} -W_{t_{k}}+W_{t_{k-1}}-W_{t_{k-1}} \right) \Delta W_{t_{k}} }}+\int_0^T W_tdW_t$
$={ \sum_{k=1}^{n}{\left( W_{t_{n}} -\Delta W_{t_{k}}-W_{t_{k-1}} \right) \Delta W_{t_{k}} }}+\int_0^T W_tdW_t$
$= W_{t_{n}}\sum_{k=1}^n{\Delta W_{t_{k}} }-\sum_{k=1}^n{\Delta W_{t_{k}}^2} -\sum_{k=1}^n W_{t_{k-1}}\Delta W_{t_{k}}+\int_0^T W_tdW_t$
$= W_{t_{n}}^2-\sum_{k=1}^n{\Delta W_{t_{k}}^2} $
$= W_{T}^2-T=2\int_0^TW_t\,dW_t$
For intuitive understanding of the non-adapted (and adapted!)integral, it helps to think of approximating the integrand by a sequence of step functions, and then multiplying the process values in each interval by the Brownian increment, and summing across the intervals.
Q2 can be rephrased as follows, and answer should follow from the above:
$$\int_0^T\int_0^TdW_s\,dW_t \neq \int_0^TdW_s \int_0^TdW_t?$$