I don’t think the truth-teller is limited to only yes or no: We are told that Random will always answer ‘da’ or ‘ja’, but we are not told that about the other two. So if we ask a yes-no question and the answer is anything other than ‘da’ or ‘ja’, we know that the god we asked is not Random, and whatever he said must be his word for ‘I don’t know’.
The solution to the puzzle follows so naturally that I won’t insult you by explaining it 
(I hope I didn’t get some minor thing backwards somewhere above, but experience guarantees I did.)
At some point, we’re going to have to move out of spoiler tags if we hope to have any real discussion. Anyway,
The information-theoretic/counting* proof of this is as follows:
Suppose that, in addition to the 6 normal role-assignment possibilities, there are also K additional role-assignment possibilities (where people still respond yes/no in some fashion or another), and your task is to determine which of the 6 + K possibilities holds.
We may assume, without loss of generality, that you always address your initial question to Person A. Now, both the normal role assignments where A is Random will be compatible with either response to your initial question. As for the other 4 + K role-assignments, each will be compatible with at least one of the responses to your initial question. Thus, some response to your initial question will be compatible with the two normal role-assignments where A is Random, plus (4 + K)/2 further role-assignments; i.e., it must be possible to end your first question with 2 + (4 + K)/2 role-assignments still in play. With each further question you ask, at least one response must be compatible with at least half of these; thus, after your two remaining questions, it must be possible to end up with >= (2 + (4 + K)/2)/2^2 = 1 + K/8 possibilities still in play. If K is positive, this is > 1, meaning there must be at least 2 possibilities still in play, meaning you can’t be sure which one holds. Thus, if there are any additional possible role-assignments at all, the task is impossible, even though the total number of possible role-assignments may remain <= 8.
*: for information theory is just logarithmic probability theory, and probability theory is just normalized measure theory, and measure theory is just weighted combinatorics, and combinatorics is just a fancy word for counting
And, incidentally, a corollary of the above tying in with a previously mentioned point,
The above counting argument also extends Chronos’s observation as far as one could have imagined: it proves that no winning strategy ever allows you to ever, even by accident in some particular situations but not others, determine the language. (For if you hoped to be able to distinguish between the two languages in the case of some particular role-assignment, then you could analyze that role assignment as two correspondingly distinct possibilities in the above argument, one “normal” and one “additional”).
It’s frankly amazing how useful the Pigeonhole Principle is, given how obvious it seems. But just giving it a name somehow seems to make it easier for humans to use it logically.
(the Pigeonhole Principle is basically that if you have more pigeons than pigeonholes, and put all pigeons into holes, there will be at least one hole that has multiple pigeons in it)