We have \(7^{2017}=\left(7^4\right)^{504}\cdot7\)

We have \(7^4\equiv1\left(mod12\right)\)

\(\Rightarrow\left(7^4\right)^{504}\equiv7^{2016}\equiv1\left(mod12\right)\)

\(\Rightarrow7^{2017}\equiv1\cdot7\equiv7\left(mod12\right)\)

So the remainder is 7.