Given in the question Three coins are tossed simultaneously.
When three coins are tossed then the outcome will be anyone of these combinations.
TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.
Thus, the total number of outcomes is $8$.
(i) now for exactly two heads, the favorable outcome are THH, HHT, HTH
thus, the total number of favorable outcomes is $3$.
We know that the Probability = Number of favorable outcomes/ Total number of outcomes
Therefore, the probability of getting exactly two heads is $3/8$
(ii) Now For getting at least two heads the favorable outcomes are HHT, HTH, HHH, and THH
Thus, the total number of favorable outcomes is $4$.
We know that the Probability = Number of favorable outcomes/ Total number of outcomes
Therefore, the probability of getting at least two heads when three coins are tossed simultaneously $=4/8=1/2$