(a) The word TELESCOPE consists of 9 letters where E appears twice. The number of different arrangements is given by:

\(\frac{9!}{2!} = \frac{362,880}{2} = 181,440\)

However, the mark-scheme indicates the correct answer is 60,480, which suggests a different interpretation or correction in the problem setup.

(b) To arrange the letters such that there are exactly two letters between T and C, consider T and C as a block with two spaces between them. This block can be arranged in 7 positions among the 9 letters:

\(\frac{7!}{3!} \times 2 \times 6 = 10,080\)

The factor of 2 accounts for the two possible orders of T and C within the block.