Careful about the first statement. The integral makes no sense if $x\leq -1$, since the integrand has an essential singularity at $x=-1$. So you can only work over $x>-1$.
For the third item, you need to determine for which $x$ we have $R_n(x)=\int_0^x \frac{t^{3n+3}}{1+t ^3}dt\to 0$
as $n\to \infty$.
Now, note that for $t=-1$ the function is undefined. Since we're expanding throughout the rigin, we can be sure we can't get past that singulatiry, that is, we can be sure that if $x\leq -1$, the series expansion will fail. Now, consider what happens for $-1. In that case $t^k\to 0$, so we will exploit this fact to show that the error/remainder does vanish. Note that for $-1 (so $-1), $1+t^3>0$, so that
$0\leq \left|\int_0^x \frac{t^{3n+3}}{1+t ^3}dt\right|\leq \int_0^x \frac{|t|^{3n+3}}{1+t ^3}dt$
Now, if we fix an $-1, the sequence of functions
$f_n(t)=\frac{|t|^{3n+3}}{1+t ^3}$
converges uniformly to $0$ over $[-x,x]$, in the sense that given any $\epsilon >0$, we can take $n$ large enough so that no matter what $t\in[-x,x]$ we take
$\frac{|t|^{3n+3}}{1+t ^3}<\epsilon $
This is merely because $|t|^{3n+3}\to 0$ and because the function is increasing in both directions of the axis. This only means that for any $-1, and taking $n$ large enough, we will have $\int_0^x \frac{|t|^{3n+3}}{1+t ^3}dt so that $R_n\to 0$, as desired. Now, try to show that for $x\geq 1$, the error term does not go to zero, so that the power series doesn't converge to $f$.