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$0^*$ will generate any number of repetitions of $0$, and will also generate the empty string $\epsilon$.
So, the problem with your expression is that $11$ is also accepted, since we can write it as $$11=(\epsilon 1\epsilon) (\epsilon 1 \epsilon)\in (0^*10^*)(0^*10^*)\subseteq(0^*10^*)^*$$
To fix this problem, use the $0^+$ operator instead, which acts just ...
1
The automaton is deterministic, so any string over $\{0,1\}$ has a unique path.
We need at least one $1$ to move from $q_1$ to the component where is the accepting state.
Immediately after reading $1$ we are always in accepting state $q_2$.
Look at the last $1$ in the input. At that moment we accept in $q_2$. To return to the accepting state, only reading $0$...
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