Why is the radius of convergence is $(-1/2, 1/2)$? 
What is the radius convergence of: $$\sum_{n=1}^\infty \frac{4^n}{n}x^{2n+3}$$

So we can take $x^3$ out of the sum: $$x^3 \sum_{n=1}^\infty \frac{4^n}{n}(x^2)^n$$
Now we substitute $y=x^2$: $$\sum_{n=1}^\infty \frac{4^n}{n}y^n$$.
Easy to see that the radius of convergence is $y=x^2\in [-\frac{1}{4}, \frac{1}{4})$ 
Now, how do I get back to $x$ properly? 
it should be $x\in(-\frac{1}{2}, \frac{1}{2})$ - Why is it open interval?

For real $x$, the number  $x^2$ is in the interval $[-1/4,1/4)$ if and only if $|x|\lt\sqrt{1/4}=1/2$.
As to why open interval, note that at $x=\pm \frac{1}{2}$ we get the harmonic series, which diverges.
Remark: The substitution $y=x^2$ is a good idea if it makes things look more familiar, but one can also work directly with the series as given, using either the Ratio Test or the Root Test.
