DFA to regular expression conversion
I'll be using the solution to $Q = \alpha Q \cup \beta$ given by $Q = \alpha^* \beta$, essentially as you would go solving a system of linear equations by hand:

$$
\begin{align*}
Q_0 &amp;= a Q_0 \cup b Q_1 \cup \epsilon \\
Q_1 &amp;= (a \cup b) Q_1 \cup \epsilon
\end{align*}
$$
From the first equation you have $Q_0 = a^* (b Q_1 \cup \epsilon)$, the second one reduces to $Q_1 = (a \cup b)^* \epsilon = (a \cup b)^*$. Replacing $Q_1$ in $Q_0$ gives:
$$
Q_0 = a^* (b (a \cup b)^* \cup \epsilon) 
    = a^* b (a \cup b)^* \cup a^*
$$