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My question reads:

Find the order of t^1000 where the permutation is ( I have converted it to disjoint cycles), (1 3 8)(2 7)(4 9 6 5).

I know the order is the lcm of the lengths of the disjoint cycles so for t alone I found the order to be 12.

I am not too sure how to do this for t^1000. Would I have to just raise 12^1000 or am I missing a step completely?

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    You know that $t$ has order 12. That means that $t^{12} = 1$ (the identity). Then $t^{1000} = t^{988}$. Do you see how to proceed? – 2017-02-10
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    @JoshuaRuiter oh...so I would have to break down the exponent to use the 12 I have that gives me 1 – 2017-02-10

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The fact that the order is $12$ says that $t^{12}$ is the identity, as is any power of $t$ that is a multiple of $12$. Then $t^{1000}=t^{1000 \mod 12}=t^4$. What power do you need to raise $t^4$ to to get the identity?

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    you would have to raise it to the third power – 2017-02-11
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    That is right. $(t^{1000})^3$ is the identity, so the order of $t^{1000}$ is $3$. Because $1000$ is a multiple of $4$, the cycles of length $2$ and $4$ are already back at the start, so they don't contribute to the order of $t^{1000}4 – 2017-02-11
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    I am a bit confused. So once I have that the order of t is 12 why did you use it as mod in the power for t^1000? – 2017-02-11
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    Because you found $t^{12}=1$, so $(t^{12})^n=t^{12n}=1$ for any $n$. This tells us the powers of $t$ are isomorphic to the cyclic group of order $12$. We keep going around and around and all we care about is the partial cycle left over at the end. – 2017-02-11
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    This would not be so easy if the power was $1001$ instead. – 2018-07-05
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More generally, if $g \in G$ has order $n$, then $g^k$ has order $n/\gcd(n,k)$.

Thus, $t^{1000}$ has order $12/\gcd(12,1000)=3$.