You are confusing the terms "independent" and "mutually exclusive". These are not the same thing. In fact events cannot be both "independent" and "mutually exclusive". It's either one, the other, or neither.
"Mutually exclusive" simply means that the two events cannot happen together. If A happens then B does not and if B happens A does not.
"Independent" simply means that the occurrence of one event is not conditional on the occurrence of the other. The probability of A happening does not depend on whether B happens or not, and vice versa.
Let $H_n$ be the indexed event of getting a head on the $n^{th}$ flip.
Given an unbiased coin, $P(H_1)=P(H_2)=\frac 1 2$
These events are independent so $P(H_1 \cap H_2) = P(H_1)\times P(H_2)$. The outcome of one coin toss does not influence the outcome of the other.
However they are not mutually exclusive, so $P(H_1 \cup H_2) = P(H_1)+P(H_2) - P(H_1 \cap H_2)$. Both coins can turn up heads.
Putting it together: $$\therefore P(H_1 \cup H_2) = \frac 12 + \frac 1 2 - \frac 12 \times \frac 12 = \frac 3 4$$
I think it is worth stressing that you can think of the outcomes as ordered or not, as you wish. That does not change the problem, nor does it change the final answer, but it does alter the computation.
Ordered: There are $4$ possible outcomes $$HH, HT, TH, TT$$ each of which has probability $\frac 14$. By inspection, the winning outcomes are $HT, TH$ so the answer is $\frac 12$.
Unordered: Now there are $3$ possible outcomes $$\{H,H\}, \{H,T\}, \{T,T\}$$
but these three are not equi-probable. Indeed, there is only one way to get $\{H,H\}$, say, which makes the probability of that outcome $\frac 14$. Same for $\{T,T\}$. There are two ways to get $\{H, T\}$ so that has twice the probability, again yielding $\frac 12$.
This comes up for dice throws, for example. If you consider ordered outcomes, then each outcome has probability $\frac 1{36}$. If you consider them as unordered, then the probability of getting distinct values, like $\{1,2\}$ is $\frac 2{36}=\frac 1{18}$ while the probability of getting a double like $\{1,1\}$ is still $\frac 1{36}$. Personally, I try to stick to equi-probable cases wherever possible, as that simplifies the computations. But it's up to you, and in some situations you really can't avoid considering outcomes with differing probabilities.
Best Answer
As you say, the probability of a person getting one head and one tail is $\frac12$. The probability of getting two heads or two tails is $\frac12$. Altogether, we'll have:
$${3 \choose 2}\times\left(\frac12\right)^2\times\frac12 = \frac38.$$
The $\left(\frac12\right)^2$ accounts for the two people who get a head and a tail, and the final $\frac12$ accounts for the one person who doesn't.