jmc

We use the property that $a\equiv b(\mathrm{mod}m)$ implies $a^c\equiv b^c(\mathrm{mod}m)$.

Since $129\equiv -3(\mathrm{mod}11)$ and $96\equiv -3(\mathrm{mod}11)$, we have $$129^{34}+96^{38}\equiv (-3{)}^{34}+(-3{)}^{38}\equiv 3^{34}+3^{38}(\mathrm{mod}11).$$Since $3^5\equiv 1(\mathrm{mod}11),$ we can see that $3^{34}=(3^5{)}^6\cdot 3^4$ and $3^{38}=(3^5{)}^7\cdot 3^3.$

Then, $$\begin{array}{rl}129^{34}+96^{38}& \equiv (3^5{)}^6\cdot 3^4+(3^5{)}^7\cdot 3^3\\ & \equiv 3^4+3^3\\ & \equiv 81+27\\ & \equiv 108\\ & \equiv \boxed{9}(\mathrm{mod}11).\end{array}$$