Help with some simpler symmetric group $S_n$ problems. I apologize if the problems seem trivial but I have not been able to find example problems or solutions to some of these questions. Could someone please confirm my attempts are correct or not?
Questions like Problems 1, 2b, 2d and 3d in particular have me confused. Why does order affect the simplification process when applying $\alpha$ or $\sigma$ multiple times?

Problem 1

My Attempt
Is $\alpha^{121}=(\alpha^{12})^{10}\alpha^1=\alpha=(125)(34)(6789)?$ Some explanation on the simplification process would be helpful.

Problem 2

My Attempt
a) I got $\sigma=(25698)(37)$ and $|\sigma|=lcm(2,5)=10$
b) Is $\sigma^{102}=(\sigma^{10})^{10}\sigma^2=\sigma^2=(26859)$? Please explain.
c) I got $\sigma=(28)(29)(26)(25)(37)$
d) I don't know. Do I reflect part of the disjoint cycle of $\sigma^2$?

Problem 3

My Attempt
a) I got $\sigma=(1496)(28537)$
b) I got $\sigma^5=(1496)$
c) $|\sigma|=lcm(4,5)=20$. I found $\sigma^2=(19)(46)(25783)$ and thus $|\sigma^2|=lcm(2,5)=10$ 
d) $\sigma^{61}=(\sigma^{20})^{3}\sigma^1=\sigma^1=(1496)(28537)$
e) I got $\sigma=(16)(19)(14)(27)(23)(25)(28)$ 
 A: Everything you've done, you've done correctly, as far as I can see. Here are a few comments:
For 1 your computation of $\alpha^{121}$ is correct. An extra step to throw in if you're feeling uncertain might be $(\alpha^{12})^{10}\alpha^1= e^{10}\alpha=\alpha$, since $\alpha^{12}$ and $e$ are the same thing ($e$ is the identity element, the trivial permutation that does nothing).
For 2b, it's the same as for 1, that $(\sigma^{10})^{10} = e^{10} = e$, and that factor can therefore be ignored.
2c is not completely answered, you need to note that $5$ is an odd number, and therefore $\sigma \notin A_9$. The product you've written out is correct.
For 2d, since you know $\sigma^{10} = e$, you know that $\sigma^8\sigma^2= \sigma^2\sigma^8 = e$, which means that the inverse of $\sigma^2$ is $\sigma^8 = (29586)$.
A: Seems like your problem is the exponentiation of permutations.
I explain the first computation $\alpha^{121}$.
First, do a integer division of the exponent by the order of $\alpha$:
$$ 121 = 10\cdot 12 + 1$$
So
$$\alpha^{121} = \alpha^{12\cdot 10 + 1}$$
Now by the exponentiation rules
$$\alpha^{12\cdot 10 + 1} = (\alpha^{12})^{10} \cdot \alpha$$
Since the order of $\alpha$ is $12$, we have $\alpha^{12} = \operatorname{id}$. (This is the answer to your central question: "Why does order affect the simplification process when applying α or σ multiple times?")
So
$$(\alpha^{12})^{10} \cdot \alpha = \operatorname{id}^{12} \cdot \alpha = \operatorname{id}\cdot \alpha = \alpha.$$
Hope this helps. You should be able now to do the remaining exponentiations in the same manner.
