Problems calculating Z-transform
We have
$$G(z)=\dfrac{z^{-1}}{1-e^{-3}z^{-1}}$$
and
$$H(z)=\dfrac{1}{1-z^{-1}}$$
and we want to calculate
$$\dfrac{C(z)}{R(z)}=\dfrac{G(z)}{1+H(z)G(z)}$$

So let's do that:
$$
\require{cancel}
\dfrac{\dfrac{z^{-1}}{1-e^{-3}z^{-1}}}{1+\dfrac{1}{1-z^{-1}}\dfrac{z^{-1}}{1-e^{-3}z^{-1}}} = \frac{z^{-1}(1-z^{-1})}{(1-z^{-1})(1-e^{-3}z^{-1}) + z^{-1}}\\
= \frac{z^{-1}(1-z^{-1})}{1 - z^{-1} - e^{-3}z^{-1} + e^{-3}z^{-2} + z^{-1}}
= \frac{z^{-1}(1-z^{-1})}{1 \cancel{- z^{-1}} - e^{-3}z^{-1} + e^{-3}z^{-2} \cancel{+ z^{-1}}}\\
= \frac{z^{-1}(1-z^{-1})}{1 - e^{-3}z^{-1} + e^{-3}z^{-2} }
$$
so it looks like either the solution you have is wrong, or that $G(z)$ is not right.



So, let's see what could be wrong. Either
$$
z \not = e^{sT}\tag{1}
$$
or
$$
s \not = 1 - z^{-1}\tag{2}
$$
or
$$
G(z) \not = \dfrac{1}{1-e^{-3}z^{-1}} \cdot {\cal T}_{s \rightarrow z}(e^{-s})\tag{3}
$$
where ${\cal T}_{s \rightarrow z}$ is the $s$ to $z$ conversion.

Comparing (1) and (2), it looks like (1) is assuming the starred transform and (2) is assuming the impulse invariance method --- which can't both be right.