You wish to prepare your state into the uniform superposition of three qubits $\{q_1,q_2,q_3\}$ such that $q_3=q_1\oplus q_2$, starting from some canonical basis element (which is conventionally the all-zeroes ket).
The following will work to prepare such a state:
See the direct Quirk link.
Generalizing this to more than three qubits, say to prepare the uniform superposition of all $n$ qubits such that $q_n=q_1\oplus q_2\oplus\cdots\oplus q_{n-1}$, one would initially perform $n-1$ Hadamard gates on the first $n-1$ qubits, then separately have these perform a controlled-$X$ (controlled-NOT) on the $n$th qubit.
As to the other question about preparing uniform superpositions for other interesting states or interesting subsets of the $2^n$ computational basis elements, there are likely too many such circuits to be formally provided, even for small $n$! There are $2^{2^n}$ different boolean functions on $n$ bits.
One approach you could always do is to prepare the uniform distribution over all $n$ qubits, calculate your function or equation of interest into an ancilla qubit, and measure the ancilla and post-select on getting the state of interest (e.g. to $|0\rangle$ or $|1\rangle$). The probability of successfully post-selecting depends on the properties of your function (e.g. how likely it is to be $|0\rangle$ or $|1\rangle$).
This would work for the "XOR state" in the original question - a fourth ancilla could store the result of the boolean test of whether $q_3=q_1\oplus q_2$, and measuring the ancilla and post-selecting upon getting $|1\rangle$ in the ancilla gives you the desired state on the other three qubits. You would get $|1\rangle$ 50% of the time. But, such a folk approach probably isn't written down anywhere formally.
