Yes, you can, but the amount of bits (that get translated into characters if you'd use, say hexadecimal encoding) that will be identical to the original hash are of course limited.
Say a hidden file in the folder contains a large counter $c$. The total hash $h_2^c$ will have a pseudo-random value. The first bit of the hash will be identical to the first bit of hash $h_1$ with a probability of $1 \over 2$. That the next bit also matches has a probability of $1 \over 4$, which is ${1 \over 2} \times {1 \over 2}$ or $2^{-2}$. If you try enough counter values you will quickly find a hash where the first few bytes are identical.
Of course the chances of finding a full MD5 hash of 128 bits this way are very close to zero. 48 bits or so may be doable (your counter $c$ should be at least 48 bits but preferably larger, say 64 bits).
If the contents of the first folder $f_1$ can have a pre-calculated value inserted then you may be able to attack MD-5 and create a fully identical hash as MD5 is rather broken. But from your description this doesn't seem to be the case.
Furthermore, if $h_1$ is set then you cannot take advantage of the birthday problem, which halves the security of the hash. It's much easier to find two hashes $h_x$ and $h_y$ that match for $n$ bits than finding a single $h_x$ that matches a preset $h_1$.