Hi Archana Sebastian,
Let 'N' be the last two digits of the number $$67^{225}$$.
We can say that N = $$67^{225}$$ mod $$100$$
$$\Rightarrow$$ N = $$(67^{224}*67)$$ mod $$100$$
$$\Rightarrow$$ N = $$((67^2)^{112}*67)$$ mod $$100$$
$$\Rightarrow$$ N = $$({4489}^{112}*67)$$ mod $$100$$
$$\Rightarrow$$ N = $$({89}^{112}*67)$$ mod $$100$$
89 when divided by 100, we can write it as -11 (In terms of the negative remainder), and since 89 power 112(even), -11 [power] 112 will be 11^112.
$$\Rightarrow$$ N = $$({11}^{112}*67)$$ mod $$100$$
$$\Rightarrow$$ N = $$(21*67)$$ mod $$100$$
$$\Rightarrow$$ N = 07
Please let me know if you have any other queries regarding the same.
I hope this helps!

Hi, 
You can consider cyclicity when considering the last digit alone, it becomes rather complex for the last two digits. In questions where you need to find the last digit or the last two digits, the right way to go about it is by finding the remainder when the number is divided by 100 or 1000 respectively. 
I hope this helps!