Calculating the pulse transfer function
You have used the correct expression for the zero-order hold, and the transfer function is indeed:
\begin{equation}
H(s) = \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)}
\end{equation}
However, it's unclear how you proceeded from here. It looks like you've simply replaced s by z (z is actually equal to esT) and then split the function into partial fractions.

Alternatively, you can first find h(t) and then find H(z):

\begin{eqnarray}
H(s) &amp;=&amp; \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \\ &amp;=&amp; (1-e^{-sT}) \left \{ \dfrac{3}{s} - \dfrac{4}{s+1} + \dfrac{1}{s+4} \right \}
\end{eqnarray}
Taking the Inverse Laplace Transform:
\begin{eqnarray}
h(t) &amp;=&amp; f(t)u(t) - f(t-T)u(t-T)
\end{eqnarray}
where 
\begin{equation}
f(t) = 3 - 4e^{-t} + e^{-4t}
\end{equation}
H(z) can now be found from h(t). You can refer to this link for a table of formulae: http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html