iterated discrete log problem
I don't know the general answer, however, it appears that Baby-step Giant-step is able to give you the solution in $O(\sqrt{in})$ time (where $n$ is the size of $G$); this is $O(\sqrt{i})$ times longer than it takes the same algorithm to solve a single discrete-log problem.

The first observation is that if you know the group order $n$ and a group generator $g$, that this problem can be reduced to $2i$ discrete log problems with the same base; if we find the values $y_a, z_a$ such that $g^{y_a} = g_a$, $g^{z_a} = h_a$, then we know $x_a = y_a^{-1} z_a \bmod n$.

So, here is how we can use Baby-Step/Giant-Step to solve all $2i$ discrete log problems in $O(\sqrt{in})$ time:


Let $t$ be the integer closest to $\sqrt{2in}$
Compute the $t$ values $g^j$, for $0 \le j &lt; t$, and store those values in a hash table.
For each $g_a, h_a$ value, compute the $n/t$ values $g_ag^{-tk}$ for $0 \le k &lt; n/t$.  Look up each value in the hash table; when we find $g_ag^{-tk} = g^j$, we know $g_a = g^{tk + j}$


The first step involves computing $t = O(\sqrt{in})$ values, and hence takes $O(\sqrt{in})$ time.  The second step involves iterating over $2i = O(i)$ variables, each iteration involving $n/t = O(\sqrt{n/i})$ time, for a total of $O(\sqrt{in})$ time.

Now, this gives an answer to the question from a theoretical standpoint ("yes, there do exist ways where we can take advantage of multiple subproblems"), however, it's not very interesting from a practical basis -- groups small enough for Baby-step/Giant-step to be a threat are too small to give real security.