We have to first find the compound function:
[tex]\begin{gathered} (m\circ n)(x)=m(n(x))=\frac{n(x)+5}{n(x)-1} \\ \frac{n(x)+5}{n(x)-1}=\frac{(x-3)+5}{(x-3)-1}=\frac{x+2}{x-4} \\ \Rightarrow(m\circ n)(x)=\frac{x+2}{x-4} \end{gathered}[/tex]The domain of this compound function is all the real values but x=4, where the function has a discontinuity.
Then, any function that has a domain of "all real values but x=4" will have the same domain as our compound function.
This is the case for h(x)=11/(x-4), because it also has one discontinuity at x=4.
Answer: Third option, h(x)=11/(x-4)
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