Write a regular expression - Contains an equal number of 01 and 10 subtrings
Part of your confusion stems from the fact that there are two different formulations of Arden's Lemma, which I'll call LR and RL. To illustrate them, let's take a simple FA with three states, $P,Q,R$, where $P$ is the start state and $R$ is the only final state. The transitions are
$$\begin{array}{c|cc}
 &amp; 0 &amp; 1\\ \hline
P &amp; Q &amp; \varnothing\\
Q &amp; R &amp; Q\\
R &amp; R &amp; R
\end{array}$$
It's clear with a moment's thought that the language accepted by this FA is denoted by the regular expression $01^*0(0+1)^*$, so let's see what the two versions of Arden's Lemma produce.

LR: In this version we generate the regular expression from left to right. If we have a state $X$ with a transition on $a$ to state $Y$ we will include the equation $X=aY$, along with any other terms that arise from transitions from $X$ to other states. Along with this, we add a "$+\epsilon$" to the expression for $X$ if $X$ happens to be a final state. Having made the collection of equations, we may apply the simplification rule that says that whenever we have $X=rX + s$ for regular expressions $r, s$ we can replace the equation by its solution $X=r^*s$. Note that here the goal is to product a regular expression for the start state.

In the example above, we'll have the equations
$$\begin{align}
P&amp;=0Q\\
Q&amp;=0R+1Q\\
R&amp;=(0+1)R+\epsilon &amp; \text{since R is final}
\end{align}$$
Substituting, we see that $R=(0+1)^*\epsilon = (0+1)*$ and $Q=1^*(0R)=1^*0(0+1)^*$, and finally $P=0Q=01^*0(0+1)^*$, as we expected.

RL: In this version we construct the regular expression from right to left. The difference is that if we have a state $X$ with transition on $a$ to state $Y$ we include the equation $Y=Xa$, along with any similar terms. In this case, we add the "$+\epsilon$" only to the start state's equation and the substitution in this case says that if we have $X=Xr+s$ we have the solution $X=sr^*$.

Returning to our example, we'll have
$$\begin{align}
P&amp;=\epsilon &amp; \text{since $P$ is the start state and has no incoming transitions}\\
Q&amp;=P0+Q1\\
R&amp;=Q0+R(0+1)
\end{align}$$
Now our goal is just opposite of what we had in the LR version: we want an expression for the final state, $R$. Working from the top down we have $Q=Q1+P0=Q1+0$ so $Q=01^*$ and $R=Q0(0+1)^*=01^*0(0+1)^*$: the same result generated in a different order.

Assuming you want to use the LR version, your equations should be
$$\begin{align}
a_1&amp;=0a_2\\
a_2&amp;=0a_3+1a_5+\epsilon\\
a_3&amp;=0a_3+1a_4+\epsilon\\
a_4&amp;=0a_3+1a_4\\
a_5&amp;=0a_6+1a_5\\
a_6&amp;=0a_6+1a_5+\epsilon
\end{align}$$
As for your problem with $a_5, a_6$, for example, we could do this:
$$\begin{align}
a_5 &amp;= 1^*(0a_6)=1^*0a_6 &amp; \text{Arden}\\
a_6 &amp;= 0^*(1a_5+\epsilon) &amp;\text{Arden again, now substitute $a_6$}\\
a_5 &amp;= 1^*0[0^*(1a_5+\epsilon)]\\
    &amp;= (1^*00^*1)a_5+1^*00^* &amp;\text{so we have}\\
a_5 &amp;= (1^*00^*1)^*1^*00^*
\end{align}$$