$$\mathcal{L}_1 = \{\text{Strings with a 1 at a multiple of 3 from the front}\} $$ and $$\mathcal{L}_2 = \{\text{Strings with a 1 at a multiple of 3 from the end}\}$$
I need to design a DFA for these.
for $\mathcal{L}_1 $ I found the regular expression $\{0\cdot \{0,1\}\cdot\{0,1\}\}^*\cdot1\cdot\{0,1\}^*$
Since if the string starts with a 1 then it doesn't have to have another 1. It is already in the set of accepted strings ($0$th position is a multiple of $3$).
Then I have the dfa:
for $\mathcal{L}_2$ I just reversed the regular expression for $\mathcal{L}_1$:
$\{0,1\}^*\cdot1\cdot\{\{0,1\}\cdot\{0,1\}\cdot0\}^*$
But I am unable to come up with a DFA to accept this regular expression, which is odd.
Edit:
So you would basically solve the language: {1 at a distance 3k and the total length is 3k +1 OR 1 at a distance 3k + 1 and the total length is 3k + 2 OR 1 at a distance 3k+2 and the total length is 3k} which is straightforward but a very large DFA: At most $12^3 = 1728$ states.