Decomposition of $H(z)$ as maximum-phase, minimum-phase
So let's check that:
$$H(z) = H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z) \tag{1}$$
where:
$$
H_{\rm min}(z) = \frac{z-1/2}{z^3}\tag{2}
$$
and
$$H_{\rm max}(z) = z-2\tag{3}$$
and
$$H_{\rm uc}(z) = z-1\tag{4}$$

Substituting $(2)$, $(3)$, and $(4)$ into $(1)$:

\begin{align}
H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z) &amp;= \frac{z-1/2}{z^3} (z-2) (z-1)\\
 &amp;= \frac{z-1/2}{z^3} \left(z^2 - 3z + 2\right)\\
&amp;= \frac{1}{z^3} \left(z^3 - 3z^2 + 2z - \frac{1}{2}\left(z^2 - 3z + 2\right)\right) \\
&amp;= \frac{1}{z^3} \left( z^3 - \frac{7}{2} z^2 + 5 z - 1\right)  \\
&amp;\not= H(z)
\end{align}

So something is wrong!

Try:
$$
H_{\rm min}(z) = 1-1/2z^{-1}\tag{2A}
$$
and
$$H_{\rm max}(z) = 1-2z^{-1}\tag{3A}$$
and
$$H_{\rm uc}(z) = 2(1- z^{-1})\tag{4A}$$

Now:
\begin{align}
H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z) &amp;= \left(1-1/2z^{-1}\right) \left(1-2z^{-1}\right) \left(1-z^{-1}\right) 2\\
&amp;= \left(1 - \frac{5}{2} z^{-1} + z^{-2}\right) \left(1-z^{-1}\right) 2\\
&amp;= 2 - 5 z^{-1} + 2z^{-2} - \left(2z^{-1} - 5 z^{-2} + 2 z^{-3}\right)\\
&amp;= 2 - 7 z^{-1} + 7 z^{-2} - 2z^{-3}\\
&amp;= H(z)
\end{align}



I believe your interpretation of Oppenheim and Schafer (OS) is incorrect. And I can find no reference that says the poles have to be outside the unit circle for a maximum phase system. This one and the one I referenced in the comments both only mention the zero locations.

And this is what my copy of OS says:



which also only mentions zero locations.

I believe exercise 5.63 is in error.  The definition used in the body of the book is:



which again does not mention pole locations.

Also, having 
$$
H_{\rm max}(z) = H_{\rm min}\left(z^{-1}\right) z^{-M_i}
$$
ensures that $H_{\rm max}$ is causal for FIR $H_{\rm min}$. 

So you have two options: either the exercise 5.69 definition is wrong, or the relationship between $H_{\rm max}$ and $H_{\rm min}$ is wrong.