Can you get two unpredictable hashes from the same message with, e.g. SHA1? I'd like to create a database of questions and answers so that an attacker with access to the database can't obtain either the questions or the answers.

If H0 and H1 are secure hashes, looking up the answer to a question might proceed as follows:

lookup(H0(question)) -&gt; value
H1(question) XOR value -&gt; answer


Can this be done securely with a hash function such as SHA1?

I've looked at HMAC, which could be used to combine the message with additional strings. The additional strings wouldn't be secret. Would access to one MAC weaken the other?

Btw.: The answer would be the same size as the hash.

EDIT

To answer a question that was asked -- the application is unique. The database would be publicly accessible, and anyone could store a question, answer pair. The encryption would be done client-side, so the server would never see plaintext. You shouldn't be able to get any useful information from seeing the database itself.
I'd need to know precisely what the "lookup" operation does but I'm pretty worried already by that XOR. Let's remove the lookup operation for now and see what happens.

Suppose we have two hashes that hash to the same length ($L$) and we XOR their outputs. You've now got a much weaker hash that either of them. 

Here's how to find a collision. Hash a message $m$ using $H_0(m) \oplus H_1(m)$ to produce an output $h$. We now wish to find a collision on $m$, such that $h=H_0(m') \oplus H_1(m')$

We now create a series of messages $m_i$ that are slightly different to $m$. The goal here is to find a message which has a xor of $H_0(m_i)$ and $H_1(m_i)$ equal to that of $H_0(m)$ and $H_1(m)$. 

Let's assume that $H_0$ and $H_1$ behave like random oracles. In that case, the probability of this happening is approximately $2^{L\over2}$. Each oracle will take the message $m_i$ and produce a random bit string of length $L$. The probability that they collide will be $2^{L\over2}$ by the birthday paradox. 

There's a slight difference here in that we're not after a precise collision but for them to have a fixed XOR. Even so, this event will happen with the same probability. We're just changing the meaning of the equality. 

Note that this attack is not the typical birthday attack against a hash. The typical birthday attack finds a pair of message $m'$ and $m$ that hash to the same value. Here, I'm taking a known fixed hash value for one message and finding collision for it. That should normally take $2^{L-1}$ time but this attack demonstrates it in $2^{L/2}$ time. Thus, this scheme is completely unsafe.