jmc

In base $3$, the place values are $\dots \dots ,3^4,3^3,3^2,3,1$. Notice that all of these except the last two are divisible by $3^2=9$. Thus, the last two digits of $a-b$ in base $3$ are the base-$3$ representation of the remainder when $a-b$ is divided by $9$. (This is just like how the last two digits of an integer in base $10$ represent its remainder when divided by $100$.)

For similar reasons, we know that $a\equiv 5(\mathrm{mod}9)$ and $b\equiv 5\cdot 6+3=33(\mathrm{mod}{6}^2)$. This last congruence tells us that $b$ is $33$ more than a multiple of $36$, but any multiple of $36$ is also a multiple of $9$, so we can conclude that $b\equiv 33\equiv 6(\mathrm{mod}9)$.

Finally, we have $$a-b\equiv 5-6\equiv -1\equiv 8(\mathrm{mod}9).$$This remainder, $8$, is written in base $3$ as $22_3$, so the last two digits of $a-b$ in base $3$ are $\boxed{22}$.