Let's suppose that we have the following $\text{GHZ}$-state
$$|\psi\rangle = \frac{1}{\sqrt{2}}|a_1a_2a_3a_4\rangle + \frac{1}{\sqrt{2}}|b_1b_2b_3b_4\rangle$$
By using quantum gates, can we evaluate $a_i\oplus b_i$, $a_{i}\oplus a_{j}$, or $b_{i}\oplus b_j$ while maintaining the superposition? Where $1\leq i, j\leq 4$.
Based on your comments in another answer, I think I see where you are going. Let your state be:
$$|\psi\rangle=\frac{1}{\sqrt 2}(|a_1a_2a_3a_4a_5\rangle+|b_1b_2b_3b_4b_5\rangle).$$
You do not wish for a circuit to prepare $|\psi\rangle$, nor do you wish to calculate the XOR of or do other controlled operations on two different qubits in your state; rather, you wish to calculate $a_i\oplus b_i$ for different $i$, while maintaining superposition. In general I don't think you can easily calculate the XOR of an individual qubit in the manner you are proposing, without violating the collision lower bound or other known lower bounds. For example, if you had an efficient way to know whether each of $a_i\oplus b_i=0$ for all qubits $i$, then you can determine that there is no such collision, as $a=b$, and hence $|\psi\rangle$ is not in a superposition of two or more states; this is what is ruled out by the collision lower bound.
But, you can find a string $d$ orthogonal to $a\oplus b$ with a quantum computer. This string might not be found easily with a classical computer. For example, let $|\psi\rangle$ be the state reached upon calculating some cryptographic 2-1 hash function of your five qubits into a second register, and measuring this second register to obtain a string $y$. Then, $a$ and $b$ are the two claws that collide at $y$. You can either (1) measure $|\psi\rangle$ to obtain the pair $(a,y)$ or $(b,y)$ but not both, or (2) perform a Hadamard transform on each of the qubits in $|\psi\rangle$ to obtain a random string $d$ such that $d\cdot (a\oplus b)=0$. This string $d$ is used in various recent proposals for proofs of quantumness, and is an interesting object in its own right.
Quantum computers are so odd and magical. They don't let you do anything you want, but they do let you do some other wonderful things.
Initially I wouldn't call your state a $W$ state, as that is a little different from what you have. You have a basis state on the first (leftmost) and fourth qubits, tensored with some version of a GHZ state on the second, third, and fifth (rightmost) qubit. Using wildcards, I would rewrite what you have as:
$$|\psi\rangle=|1.. 0.\rangle\otimes\frac{1}{\sqrt 2}\big(|.00.1\rangle+|.11.0\rangle\big)$$
Also if you perform a projective measurement on your qubits then you are left with classical information (e.g., 10001 or 11100), from which you can do what you wish.
But perhaps you wish to maintain coherence of your qubits. Here is a Quirk circuit which you might want to play with - this prepares your state. Note the endianness used here is such that the bottom qubit in the circuit is your leftmost qubit.
In particular, you can follow @Rammus's suggestion and add as many controlled gates as you'd like to your circuit. Indeed, calling your gate the SG (SuperGravity) gate, you can have some more fun with Quirk, adding random controlled NOT and controlled SWAP gates, as follows:
