What is the easiest way to determine the accepted language of a deterministic finite automaton (DFA)?

This is a very small DFA, so I’d use a fairly ad hoc approach. Notice first that its language is precisely the set of words that reach $q_2$: once it gets to $q_2$, it stays there, and there is no other acceptor state. Thus, we need only see what words get to $q_2$.

Clearly we need a $1$ at some point. It can come right away, thanks to the $q_0\overset{1}\longrightarrow q_2$ transition, or it can come after some $0$s. How many $0$s? One $0$ gets us to $q_1$, at which point a $1$ takes us to $q_2$. Two $0$s take us back to $q_0$, at which point a $1$ takes us to $q_2$. In fact, it appears that any number of $0$s (including none at all) followed by a $1$ will get us to $q_2$. Thus, anything conforming to the regular expression $0^*1$ gets us to $q_2$. After that we can have anything at all, so the language of the DFA must be $0^*1(0+1)^*$.

We can start analyzing from the final state $q_2$: $$ u(0|1)^* $$ where $u$ is a prefix word. Arrival only via accepting a $1$ coming from $q_1$ or $q_0$: $$ v1(0|1)^* $$ where $v$ is another prefix word. Joining the start state $q_0$ via the upper (via $q_1$) or lower edge (direct): $$ (00)^*(0|\epsilon)1(0|1)^* $$ which simplifies to $$ 0^*1(0|1)^* $$