(i) Consider the three Os as a single unit and the two Ts as another unit. This reduces the problem to arranging the units: \(\{OOO, TT, A, D, S, L\}\), which is 6 units in total. The number of arrangements is \(6! = 720\).

(ii) First, calculate the total number of arrangements of the letters in TOADSTOOL. There are 9 letters with 2 Ts and 3 Os, so the total arrangements are \(\frac{9!}{2!3!} = 30,240\). Next, calculate the number of arrangements where the Ts are together. Treat the Ts as a single unit, reducing the problem to arranging \(\{TT, O, O, O, A, D, S, L\}\), which is 8 units. The number of arrangements is \(\frac{8!}{3!} = 6,720\). Therefore, the number of arrangements where the Ts are not together is \(30,240 - 6,720 = 23,520\).

(iii) To find the probability that a randomly chosen arrangement has a T at the beginning and a T at the end, consider the remaining 7 letters \(\{O, O, O, A, D, S, L\}\). The number of arrangements of these letters is \(\frac{7!}{3!} = 840\). The probability is then \(\frac{840}{30,240} = \frac{1}{36}\) or 0.0278.