You shall find a formula that has the following truth table:
$$\begin{array}{cc|c}
A & B & ? \\
\hline
T & T & T \\
T & F & F \\
F & T & F \\
F & F & T \\
\end{array}$$
Reason: If the person you ask is actually a truth teller (i.e. $A$ is true), then the desired formula shall have the same truth value as B. If the person you ask is a liar (i.e. $A$ is false), though, the desired formula shall have flipped truth values, because (so to speak) the liar will flip the truth value again.
Now the desired formula is easy to come up with; it's simply $A \leftrightarrow B$.
Explanation: Ask the local "Is the following the case: you're a truth teller iff the left hand branch leads to the capital?" (assuming he will understand "iff" truth-functionally)
First assume he's a truth teller. If the left hand branch leads to the capital, the biconditional is true and he will answer "yes". If the left hand branch does not lead to the capital, the biconditional is false (since the two sentences differ in truth value), and he will answer "no".
Second assume he's a liar. If the left hand branch leads to the capital, the truth values of $A$ and $B$ differ, so the biconditional is false, and because you're asking a liar by assumption, he will answer "yes". If the left hand branch does not lead to the capital, then the biconditional is true and the answer will be "no".
In both cases, the answer is "yes" if the left hand branch leads to the capital, but "no" if it does not $-$ which is exactly the desired result, I take it.
Best Answer
This is famously called by George Boolos the "Hardest logic puzzle ever" (1996).
He, that, as far as I know, was the first to present and solve it, states the puzzle as follows:
Some hints:
Since either A, B or C is the True, False and Random, we have $3!=6$ possible cases.
Now consider those three questions:
You should ask them to no matter who, in the structure:
Note that the above indirect sentence contains two questions one inside another. The result is that even if the False lies in the innermost question, he will have to lie about lying in the out-most question and this amounts to the cancelling of the negations.
In a normal scenario, that is, where the foreign language problem is not put, then those three questions are enough to find out the identity of A, B, C.
To see this, simply construct a truth table of $2^3=8$ lines, corresponding to the possible combinations of answer to those questions: $$\begin{array} {|c|} \hline Q_1 & Q_2 & Q_3 & Interpretation\\ \hline \color{red}{1} & \color{red}{1} & \color{red}{1} & --\\ \hline 1 & 1 & 0 & \text{A: Random, B: True, C: False} \\ \hline 1 & 0 & 1 & \text{A: True, B: False, C: Random}\\ \hline 1 & 0 & 0 & \text{A: Random, B: False, C: True}\\ \hline 0 & 1 & 1 & \text{A: False, B: Random, C: True}\\ \hline 0 & 1 & 0 & \text{A: True, B: Random, C: False}\\ \hline 0 & 0 & 1 & \text{A: False, B: True, C: Random}\\ \hline \color{red}{0} & \color{red}{0} & \color{red}{0} & -- \\ \hline \end{array}$$
Where:
$$Q_1 \equiv Random(A) \vee False(B)$$ $$Q_2 \equiv Random(B) \vee False(C)$$ $$Q_3 \equiv Random(C) \vee False(A)$$
Indeed, since there is one and only one True, False and Random, it is easy to see that the first and last lines of the table are impossible (show it!). Hence, of of it amounts to, again, 6 possible interpretations, where in every case we can find the True, False and Random's identity.
The above solution assumes we know the language of the True, False and Random.
Now, in its own language variation, we can try to overcome the fact we don't know what ja or da means by considering the following question, as firstly observed by Roberts (2001):
In fact, this answer to this question will come out as ja if the truthful answer to Q is yes, and as da if the truthful answer to Q is no. Now, what about applying this reasoning with the above solution?