Is $\sqrt[4]{-1}=\sqrt(i)$? I get $\sqrt[4]{-1}=((-1)^{\frac{1}{2}})^{\frac{1}{2}}=\sqrt{i}$.

But on the one hand $\sqrt[4]{-1}$ has $4$ roots and on the other $\sqrt{i}$ has just $2$.
The function $\sqrt[n]{x}$ is usually defined for non-negative, real $x$ and returns the non-negative, real solution to $y^n=x$. It can be extended to other domains, but must be accompanied by clear restrictions on which roots are to be returned.



For example, we can define $\sqrt[4]{z}$ for $z\in\mathbb{C}$, where the root in the first quadrant (excluding the imaginary axis) is returned.

Similarly, we can define $\sqrt{z}$ for $z\in\mathbb{C}$, where the root in the first or second quadrant (excluding the negative real axis) is returned.

With these definitions, we get $\sqrt[4]{-1}=\sqrt{i\,}=\frac{1+i}{\sqrt2}$.



However, if we consider $\sqrt[4]{z}$ to be the set valued function $\left\{w\in\mathbb{C}:w^4=z\right\}$, and similarly define $\sqrt{z}$ to be the set $\left\{w\in\mathbb{C}:w^2=z\right\}$, then $\sqrt[4]{-1}$ has $4$ elements and $\sqrt{i\,}$ has only two elements, so they are not the same, but with this definition,
$$
\sqrt{i\,}\subset\sqrt[4]{-1}
$$
Indeed as you observed $\sqrt[4]{-1}$ has four roots while $\sqrt i$ has only 2 roots thus they are not the same.

What is true is that they have roots in common.