non-deterministic finite automaton state transition table language question
Starting from the $M$ you gave in a comment (which you should really include in the question), we have the transition table
$$\begin{array}{r|cc}
\delta &amp; 0  &amp; 1 \\ \hline
q_0    &amp; q_1 &amp; q_2\\
q_1    &amp; q_0 &amp; q_2\\
q_2    &amp; q_2 &amp; q_2
\end{array}$$
with $q_0$ the start state and $q_1$ the final state. Now we modify this to a new NFA $M'$ by adding another final state $q'_0$ and construct the transitions $\delta'$:


$\delta'(q'_0, 0) = \{\delta(q_0, 0), q_0\}=\{q_1,q_0\}$, since $\delta(q_0,0)=q_1\in F$
$\delta'(q'_0, 1) = \{\delta(q_0, 1)\}=\{q_2\}$, since $\delta(q_0,1)=q_2\notin F$
$\delta'(q_0, 0) = \{\delta(q_0, 0), q_0\}=\{q_1,q_0\}$, since $\delta(q_0,0)=q_1\in F$
$\delta'(q_0, 1) = \{\delta(q_0, 1)\}=\{q_2\}$, since $\delta(q_0,1)=q_2\notin F$
$\delta'(q_1, 0) = \{\delta(q_1, 0)\}=\{q_0\}$, since $\delta(q_1,0)=q_0\notin F$
$\delta'(q_1, 1) = \{\delta(q_1, 1)\}=\{q_2\}$, since $\delta(q_1,0)=q_2\notin F$
$\delta'(q_2, 0) = \{\delta(q_2, 0)\}=\{q_2\}$, since $\delta(q_2,0)=q_2\notin F$
$\delta'(q_2, 1) = \{\delta(q_2, 1)\}=\{q_2\}$, since $\delta(q_2,0)=q_2\notin F$


giving us this transition table for $M'$
$$\begin{array}{l|cc}
\delta'     &amp; 0           &amp; 1 \\ \hline
\{q'_0\}    &amp; \{q_0,q_1\} &amp; \{q_2\}\\
\{q_0\}     &amp; \{q_0,q_1\} &amp; \{q_2\}\\
\{q_1\}     &amp; \{q_0\}     &amp; \{q_2\}\\
\{q_2\}     &amp; \{q_2\}     &amp; \{q_2\}\\
\{q_0,q_1\} &amp; \{q_0,q_1\} &amp; \{q_2\}
\end{array}$$
with $\{q'_0\}$ the start state and $\{q'_0\}, \{q_1\}, \{q_0,q_1\}$ the final states. Notice that $\{q_0\}$ and $\{q_1\}$ are unreachable so can be eliminated from the NFA. Notice also that in both FAs, state $q_2$ is a "trap" state: once you enter it (on input $1$), you'll never leave. This means that neither FA will accept any string containing a $1$ and from there it's not too hard to see what $L(M)$ and $L(M')$ are.