jmc

$h^{-1}(5)$ is defined as the number $y$ such that $h(y)=5$. Thus, we solve the equation $$\frac{1+y}{2-y}=5.$$Multiplying both sides by $2-y$, we have $$1+y=5(2-y).$$Expanding gives $$1+y=10-5y,$$then adding $5y-1$ to both sides gives $$6y=9.$$Finally, we divide both sides by $6$ and simplify to get $y=\boxed{\frac{3}{2}}$.

Note that we can check our work by plugging $\frac{3}{2}$ into the formula for $h$: $$\frac{1+\frac{3}{2}}{2-\frac{3}{2}}=\frac{\left(\frac{5}{2}\right)}{\left(\frac{1}{2}\right)}=5,$$which is what we expected.