Total number of ways in which $5$ teams can finish the race is $n(S) = 5! = 120$.
(i) $A, B, C$ finish $1^{\text{st}}, 2^{\text{nd}}, 3^{\text{rd}}$ respectively:
Positions $1, 2, 3$ are fixed for $A, B, C$ ($1$ way).
The remaining $2$ teams ($D, E$) can finish in the remaining $2$ positions in $2! = 2$ ways.
Favourable outcomes $= 1 \times 2 = 2$.
Probability $= \frac{2}{120} = \frac{1}{60}$.
(ii) $A, B, C$ finish in the first three positions in any order:
Teams $A, B, C$ can be arranged among the top three spots in $3! = 6$ ways.
Teams $D, E$ take the remaining two spots in $2! = 2$ ways.
Favourable outcomes $= 3! \times 2! = 6 \times 2 = 12$.
Probability $= \frac{12}{120} = \frac{1}{10}$.
(i) $\frac{1}{60}$, (ii) $\frac{1}{10}$