Examples of Regular, Context-free and Context-sensitive languages
$L_1$, $L_2$ and $L_4$ are correct (notice that you are not showing that $L_2$ is not regular and $L_4$ is not context-free, which can be done using the appropriate version of the pumping lemma).

As far as $L_3$ is concerned, let  $x$ be the number that is encoded in binary and call $x_i$ its $i$-th least significant bit, indexed from $0$. Then, counting modulo 5, we have:

$\displaystyle x = \sum_{i=0}^\infty x_i 2^i = \sum_{i=0,4,8..} x_i + 2 \sum_{i=1,5,9..} x_i + 4 \sum_{i=2,6,10..} x_i + 3 \sum_{i=3,7,11..} x_i$.

I.e., the least significant binary digit contributes $1$ to $x \bmod 5$, the next one contributes 2, then 4, 3, 1, 2, 4, 3, ... and this pattern repeats.

This means that you can create a DFA that recognizes $L_3^R$ (and hence $L_3$) using at most 20 states, namely $\langle i,j \rangle$ where $i \in \{0,1,2,3\}$ is the index, modulo 4, of the next character in the string, and $j \in \{0,1,2,3,4\}$ is the partial sum, modulo $5$, of the contributions of binary digits seen so far.

If the next bit is $0$ then:
$\langle i,j \rangle \to \langle i+1 \pmod 5, j \rangle$.

If the next bit is $1$ then $\langle i,j \rangle \to \langle  i+1 \pmod 5, j+ 2^i  \pmod 5\rangle$. Writing this down explicitly:


$\langle 0,j \rangle \to \langle 1, j+1  \pmod 5\rangle$;
$\langle 1,j \rangle \to \langle 2, j+2  \pmod 5\rangle$;
$\langle 2,j \rangle \to \langle 3, j+4  \pmod 5\rangle$;
$\langle 3,j \rangle \to \langle 0, j+3  \pmod 5\rangle$.


The initial state is $\langle 0, 0 \rangle$ and the final states are all the ones of the form $\langle i, 0 \rangle$.