Let the alphabet = {a, b} for $L_{1}$ and $L_{2}$
$L_{1}$ = { w | w contains an even number of a's }
$L_{2}$ = {w | w has at least 1 b after each a }
$\rule{10cm}{0.4pt}$
$\hspace{4cm}L_{1}$
$\rule{10cm}{0.4pt}$
$\hspace{4cm}L_{2}$
$\rule{10cm}{0.4pt}$
How can I find the concatenation of $L_{1}$.$L_{2}$ represented as a DFA?
You have to add two new arrows starting in the final state of the first automaton simulating what happens in the initial state of the second automaton.
I think it would be like this. You can then determinise this automaton. I am sorry it is handwritten.