Trusting Issues

Let's say that the answer was $\textbf{NOT}$ the same for the second question. You pick a tourist and they answer different in 2 different orders (ways):

$\frac2{3}\times (\frac1{4}\frac3{4}+\frac3{4}\frac1{4}) = \frac1{4}\text{ chance}$

Now the probability of getting the same answer in a row is $1-\frac3{4}=\frac1{4}$ . Getting the same correct answer in a row can only occur from the tourist with probability:

$\frac2{3}\times (\frac3{4}\,\frac3{4})=\frac3{8}$

This is exactly half of $\frac3{4}$ (the chances of getting 2 in a row, which we know happened) so the answer is $\boxed{\frac1{2}}$

If the correct answer is East, only the tourists will give you the right answer twice, and the probability of that will be $\frac34\times\frac34=\frac9{16}$ . And since the tourists are $\frac23$ of the population, the overall probability of the East/East answer will be $\frac23\times\frac9{16}=\frac38$ .

If the correct answer is actually West, the locals will give you $\frac13$ probability of East/East answer right there, and the tourist will contribute $\frac23\times\frac1{16}=\frac1{24}$ . Adding these together will come up with $\frac13+\frac1{24}=\frac8{24}=\frac38$ , exactly the same as what we got for East.

The probability of East being the correct answer is therefore $\boxed{\frac12}$