To find the last two digits of a number in general, we need to take the remainder when the number is divided by $100.$
For this particular case, we can use modular arithmetic and the concept of cyclicity of the last two digits of powers of $7.$
The last two digits of the powers of $7$ repeat in a cycle of $4: 07,49,43,01,07,49, 43,01,$ and so on.
Therefore, we can find the last two digits of $7^{45}$ by finding the remainder when $45$ is divided by $4,$ which is $1.$
Thus, the last two digits of $7^{45}$ are the same as the last two digits of $7^1,$ which is $07.$
Therefore, the answer is option A) $07.$
