A man speaks truth 3 times out of 5.He says that in tossing 6 coins,two tails appear.What is the probability that this event is actually true?
Let $E_1$ be the event he speaks truth and $E_2$ be the event he speaks a lie.
$P(E_1)=\frac{3}{5},P(E_2)=\frac{2}{5}$
Let $A$ be the event that $2$ tails appear.
We need to find $P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$
$P(A/E_1)=6C2\times(\frac{1}{2})^6=\frac{15}{64}$,What will be $P(A/E_2)?$
The answer for probability that this event is actually true is $\frac{115}{288}$
The application of Bayes's Theorem in the question is incorrect. The conditional probability $P(E_1\mid A)$ is the probability that the man says two tails, given that there were two tails. What we want is the probability that there are two tails given that the man said there were two tails.
I find it next to useless to give names to events such as "the man lies" in problems like this. In fact, assuming the man lies $\frac35$ of the time regardless of what actually happens, $P(A\mid E_1) = P(A\mid E_2) = P(A)$. And without that assumption, it is possible that the man always tells the truth when there are two tails in six coin tosses, and lies in more than $\frac35$ of all other cases to come to a total probability of $\frac35$ to lie, so without further information we could not conclude anything at all.
A better formulation is to define $B$ as the event that the man says there are two tails. Then the answer is $P(A\mid B)$ where $$ P(A\mid B) = \frac{P(B\cap A)}{P(B\cap A) + P(B\cap A^\complement)} = \frac{P(B\mid A)P(A)}{P(B\mid A)P(A) + P(B\mid A^\complement)P(A^\complement)}. $$
We still have to assume the man's decision to lie is independent of what happened, so $P(B\mid A) = \frac35.$ As you already found, $P(A) = \frac{15}{64},$ so $P(A^\complement) = \frac{49}{64}.$
We just need to know the value of $P(B\mid A^\complement)$ in order to apply the formula. That's where creative interpretation of the problem statement comes in. If the actual train of events was six coins were tossed, we asked the man if there were exactly two tails, and he said "yes," then $P(B\mid A^\complement) = \frac25.$ If instead we had asked the man how many tails occurred and he said "two," $P(B\mid A^\complement)$ depends on the answers to questions such as, "How likely is it that the man would have said 'four' rather than 'two' if there had been exactly three tails and he lied?" And if the man said there were two tails without any prompting, how do we know the probability that he would have said that particular lie in the event of three tails?
I believe the usual interpretation of problems like this is the first one: the man is (at least in effect) simply answering "yes" or "no" to the question whether there were two tails; "two" is the only possible lie he can tell when there are more or fewer than two tails. I also think this is a stupid way to frame a problem; if that is the intent, then the writer should state explicitly in the problem statement that the man is answering a yes/no question.