I know that $L(f(t-a).u(t-a)) = e^{-as}.F(s)$
Does that mean the laplace transform of the above equation is $e^{-2πs}$?
Edited: sorry i meant $(\delta(t−π/6))u(t-π/6)$ not $(\delta(t−π/6))u(t-π)$ missed the 6 out when i initially asked the question.
It is equal to $0$ since $$ f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0) $$ and $$ u({\pi\over 6}-\pi)=0 $$
Also the statement you wrote is wrong. We have $$ L(f(t-a))=e^{-as}F(s) $$ as long as the ROCs are treated carefully.