$\mathrm{Hash}(A\|B)$ where $B$ is known, $A$ is unknown. Would knowing $\mathrm{Hash}(A\|C)$ as well as $C$, speed up a brute force search for $A$ Let's say $\mathrm{Hash}(A\|B)$ is a hash of two concatenated strings, $A$ and $B$.
Suppose I am trying to figure out $A$ from $\mathrm{Hash}(A\|B)$, I know $B$ but not $A$, I want to brute force to find $A$. Let's ignore practical reality for a moment.
If I also know $\mathrm{Hash}(A\|C)$, $\mathrm{Hash}(A\|D)$ etc, as well as $C$ and $D$ etc. can I use this info to reduce the time required in the brute force search?
The answer depends on your definition of a hash function.
In general hash functions don't guarantee, that they hide their input.
A general hash function gives you three properties:

preimage resistance (given $y = H(x)$ it is hard to find $x^*$ such that $H(x^*) = H(x)$)
second preimage resistance (given $y = H(x)$ it is hard to find $x^*$ such that $H(x^*) = H(x)$ and $x^* \neq x$)
collision resistance (it is hard to find $x_1$ and $x_2$ such that $x_1 \neq x_2$ and $H(x_1) = H(x_2)$)

None of these properties guarantee that the hash function doesn't reveal part of the input - i.e. there are hash functions that reveal $A$.
However, in practice you typically model hash functions as random oracles - for any input, the function yields a randomly chosen output.
In this case it is easy to show that it is hard to find $A$.
You might also want to take a look at length extension attacks, which are a way many hash functions deviate from a random oracle. However, these won't help you in finding $A$, they will however help you in finding (roughly) $H(A \| B \| C)$, given $H(A \| B)$ (without knowing $A$ or $B$).