For the case where the heavy and light offset exactly, you can solve this in $5$ steps. This is clearly the minimum since there are $12\times 11 = 132 > 81 = 3^4$ possibilities.
Label the balls $a$ through $l$. Weigh $a,b,c,d$ vs. $e,f,g,h$. There are 48 cases where the scale will tip left, 48 where it tips right, and 36 where it stays even.
Either the heavy one is in $a,b,c,d$ or the light one is in $e,f,g,h$ or both. For step 2, weigh $a,b$ vs. $c,d$. If the scale tips left, then one of $a,b$ is heavy (16 cases), tips right, then one of $c,d$ is heavy (16 cases), stays even, then one of $i,j,k,l$ is heavy and one of $e,f,g,h$ is light (16 cases).
If it tipped left, then for Step 3 weigh $a$ vs. $b$, telling you which is heavy. For Step 4, weigh $e,f,g$ vs. $h,i,j$. If it tips, then one of those three is light and you can figure out which in one more weighing. If Step 4 came out flat, then one of $k,l$ is light, and you can determine which in one more weighing.
If Step 2 tipped right, then for Step 3 weigh $c$ vs. $d$ and after that proceed as above.
If Step 2 came out flat, then for Step 3 weigh $e,f,i$ vs. $g,h,j$. If it tips left, then one of $g,h$ is light, and one of $i,k,l$ is heavy, which you can finish in two more weighings. Similarly, if Step 3 tips right, then one of $e,f$ is heavy, and one of $j,k,l$ is light.
If Step 3 came out flat, then the heavy/light combination is one of $i/e, i/f, j/g, j/h$. For Step 4 weigh $i$ vs. $j$ to figure out which is heavy, then either weigh $e$ vs. $f$ or $g$ vs. $h$.
Tip Right: If Step 1 tipped right, then proceed as above, switching the labels around.
Stays Flat: If Step 1 came out flat, then one of the four sets $(a,b,c,d)$, $(e,f,g,h)$, or $(i,j,k,l)$ contains both the heavy and light balls.
For Step 2, weigh $a,e,i$ vs. $b,f,j$. Tip left indicates one of $a,e,i$ is heavy and/or one of $b,f,j$ is light: 15 possibilities. Likewise 15 possibilities to tip right, leaving 6 possibilities Step 2 stays flat.
Tip left: For Step 3 weigh $a,e$ vs. $i,c$. If it tips left, then either $a$ or $e$ is heavy ($c$ cannot be light unless $a$ is also heavy, otherwise Step 2 would not have tipped left). Compare them in Step 4 to determine which is heavy, then weigh the other three from that set in Step 4 to find the light one. Likewise, if it tips right, then one of $i$ or $c$ must be heavy and proceed as above.
If Step 3 remained flat, then none of $a,e,i$ is heavy, and one of $b,f,j$ must be light. Weight $b$ vs. $f$ to determine the light one (if they come out the same, the $j$ is the light one). Then once you know the light one, weigh the other two from that set (not $a, e, $ or $i$), and you are done.
If Step 2 came out flat, then one of the sets $c,d$, $g,h$, or $k,l$ contains both the heavy and light balls. Differentiating between these six possibilities can easily be done by weighing each pair separately, yielding in answer in at most 5 weighings.
Thus, it is always possible to solve the first problem with at most $5$ weighings.
Not perfectly offset: The heavy and light balls together could weigh more than, less than, or the same as two regular balls. This ups the possibilities to $3\times 11 \times 12 = 396$ possibilities, which would thus require at least $6$ weighings to differentiate. On the other hand, we don't need to know whether the combination is heavier or lighter--just which two they are.
The overall strategy is to examine the 396 cases and see what weighings divide them most evenly. So start with weighing $a,b,c$ vs. $d,e,f$, which splits the groups into 147 tip left, 147 tip right, and 102 stay flat. This is much more envenly divided than weighing 4 vs. 4, which yields $168, 168, 60$.
If that tipped left, you might weight $a,b$ vs. $c,g$, yielding a split of $58, 38, 51$. Following this strategy of examining the splits based on different weighings, and keeping the three groups as close together in size as possible, it should be possible to complete in $6$ weighings, but I haven't explored the depth of the problem.