How to calculate a permutation to the $-100$th power? I have a this permutation element $(1,2)(3,4,6,5,8)(7,9,10)$ and I need find the $-100$th power of this element.
I think firstly we need find $-1$ power, and I have this element: $(1,2)(3,8,5,6,4)(7,10,9)$.
Here $(1,2)$ have order $2$, so after $100$ permutation we have $(1 \to 1)$, $(2 \to 2)$ and we don't need write this. The same goes with $(3,8,5,6,4)$. But $(7,10,9)$ changes to $(7,10,9)$ in power $1$, that is $(7,10,9)$ doesn't change.
The answer is $(7,10,9)$.
Are my calculations correct?

Your calculations are fine.
An alternative approach would be to notice that the order of the element is the least common multiple of the orders of its disjoint cyclic components, so the order is ${\rm lcm}(2,5,3)=30$. Now note that
$$\begin{align}
((1,2)(3,4,6,5,8)(7,9,10))^{-100}&=((1,2)(3,4,6,5,8)(7,9,10))^{-90}\\
&\times((1,2)(3,4,6,5,8)(7,9,10))^{-10}\\
&={\rm id}\times((1,2)(3,4,6,5,8)(7,9,10))^{-10}\\
&=(1,2)^{-10}(3,4,6,5,8)^{-10}(7,9,10)^{-10}\tag{1}\\
&={\rm id}\times(7,9,10)^{-9}(7,9,10)^{-1}\\
&={\rm id}\times(7,9,10)^{-1}\\
&=(7,10,9),
\end{align}$$
where $(1)$ holds because disjoint cycles commute.
