Weblink to $1$st question and its solution :-
https://www.toppr.com/ask/question/a-speaks-the-truth-3-times-out-of-4-and-b-speaks-the-truth-2/
Weblink to $2$nd question and its solution :-
https://www.toppr.com/ask/question/a-man-is-known-to-speak-the-truth-3-out-of-4-times-he-throws-3/
My confusion is that in the first question we make an extra $\frac{1}{5}*\frac{1}{5}$ multiplication in the second case of total probability (where $2$ didn't show up on the die but still both $A$ and $B$ assert that it was a $2$) to account for the total number of ways of making the lie that $2$ came up but it didn't ( $\frac{1}{5}$ for each person)
Now in second question, in second case of total probability (where $6$ didn't show up on the die but still the man says it is a $6$). Now in this case we do not make that multiplication to account for the total number of ways of speaking the lie that $2$ didn't come up but he still says it is a $2$. Why we didn't make a $\frac{1}{5}$ multiplication here?
Please explain with other situational examples if possible, I am weak in probability. Thanks.
You're right. Something is wrong with their solution or they are making additional or different assumptions than are common for the problem.
Let us first make a few assumptions. The man chooses to lie or tell the truth independently of the result of the die. If the man lies, he will uniformly select at random a number from the numbers on the die that didn't appear and his lie to us is that the die shows the selected number. If the man tells the truth, his truth he tells is the number showing on the die. The die is standard and fair.
These match with the assumptions seemingly made in the first linked problem (which has the additional hypothesis that the two people being observed make their decisions independently)
Letting $A$ be the event a six was rolled, $B$ the event he lies, $C$ the event that he says it was a six (whether truthfully or a lie), we are asked to find $\Pr(A\mid C)$, that is... the probability that a six was actually rolled given that we were told that it was a six.
We have $\Pr(A\mid C)=\dfrac{\Pr(A\cap C)}{\Pr(C)}=\dfrac{\Pr(C\mid A)\Pr(A)}{\Pr(C)}=\dfrac{\Pr(C\mid A)\Pr(A)}{\Pr(C\cap A)+\Pr(C\cap A^c)}$
Now, the numerator we can see as being $\frac{3}{4}\times \frac{1}{6}$ as the only way he says it is a six given that it actually is a six is by telling the truth.
The denominator's first term is the same as the numerator. The denominator's second term however should expand as $\Pr(C\cap A^c) = \Pr(A^c)\Pr(B\mid A^c)\Pr(C\mid B\cap A^c) = \frac{5}{6}\times\frac{1}{4}\color{red}{\times\frac{1}{5}}$.
The linked answer makes the seemingly invalid assumption that $\Pr(C\mid B\cap A^c)=1$, that if the die did not show a six and the man chose to lie that he always says that it is a six.
With the assumptions made above, this results in a final probability of $\dfrac{3}{4}$ that he was telling the truth about what he saw on the die when he said it was a six.
Now, to be fair, perhaps the original question writer intended the man to answer only yes or no questions. We the observer asked the man the yes or no question "Was a six rolled" to which if it was and he was telling the truth he would say "yes," or if it wasn't and he were lying he would say "yes." Otherwise he would have simply said "no."
The end result is that questions like this should be incredibly specific about the actions and decision making process behind what information is shared.