I started with Fermat's Little Theorem taking $2^{12}\equiv1\pmod{13}$ and raising it to power $5$, we got $2^{60}\equiv1\pmod{13}$, then multiplied the congruence $2^{10}\equiv(-3)\pmod{13}$.
I did the same with $5$ and I get $2^{70}+5^{70}\equiv(-4)\pmod{13}$.
Is another method to find it
You're doing quite well: $2^{12}\equiv1\pmod{13}$, so $$ 2^{70}=(2^{12})^5\cdot2^{10}\equiv2^{10}\pmod{13} $$ Now note that $2\cdot7\equiv1\pmod{13}$, so $2^{10}\equiv2^{12}\cdot7^{2}\equiv7^2\pmod{13}$ and finally $$ 2^{70}\equiv49\equiv10\pmod{13} $$ With similar computations, noting that $5\cdot8\equiv1\pmod{13}$ we have $$ 5^{70}\equiv(5^{12})^6\cdot8^2\equiv64\equiv12\pmod{13} $$ This shows $$ 2^{70}+5^{70}\equiv10+12\equiv9\not\equiv0\pmod{13} $$