Brute-force attack given multiple hash prefixes
If you have two prefixes, say $p_1$ and $p_2$ assuming a well designed hash function, this will give you two lists of possible candidates $L_1,L_2$ each of size roughly $2^{24}.$ The correct value is in both of these lists. It is unlikely to also have spurious candidates since assuming uniformity o relevant variables and fixing, say the list $L_1$, the probability that a random quantity from $\{0,1\}^{56}$ is not in $L_1$ is 
$$
\left(1-\frac{2^{24}}{2^{56}}\right)=\left(1-\frac{1}{2^{32}}\right),
$$
since $\lim_{n\rightarrow \infty}\left(1+\frac{x}{n}\right)^n=\exp(x).$ Thus
the probability that the $2^{24}$ elements in $L_2$ all fall outside $L_1$ is
$$
\left(1-\frac{1}{2^{32}}\right)^{2^{24}}=\left[\left(1-\frac{1}{2^{32}}\right)^{2^{32}}\right]^{2^{-8}}\approx \exp(-2^{-8}) \approx 0.9961
$$
so only in one of about 250 trials would there be a spurious candidate in addition to the correct candidate.

Edit: It is widely believed and supported by experimentation that SHA-1 is such a function. Ignore my earlier reference to universal hash functions.