There are two mistakes in your calculations. The first one has been pointed out in Ahmad Bazzi's answer. However, I would suggest a different expansion of the given function $X(z)$, which requires the introduction of only two unknown constants:
$$X(z)=2.5+\frac{Az^{-1}}{1-1.5z^{-1}}+\frac{Bz^{-1}}{1-0.8z^{-1}}\tag{1}$$
I'm sure you can determine these two constants yourself.
The second mistake is the inverse transform of the basic function
$$G(z)=\frac{1}{1-\alpha z^{-1}}\tag{2}$$
For $|z|>\alpha$ the inverse $\mathcal{Z}$-transform of $(2)$ is
$$g[n]=\alpha^nu[n]\tag{3}$$
where $u[n]$ is the unit step sequence.
For $|z|<\alpha$, the inverse $\mathcal{Z}$-transform of $(2)$ is
$$g[n]=-\alpha^nu[-n-1]\tag{4}$$
From $(2)-(4)$, the inverse $\mathcal{Z}$-transform of $(1)$ is
$$x[n]=2.5\delta[n]-A(1.5)^{n-1}u[-n]+B(0.8)^{n-1}u[n-1]\tag{5}$$
For $n=0$ you have contributions from two terms:
$$x[0]=2.5-A(1.5)^{-1}\tag{6}$$
Note that another possible expansion of $X(z)$ is
$$X(z)=1+\frac{C}{1-1.5z^{-1}}+\frac{D}{1-0.8z^{-1}}$$
resulting in
$$x[n]=\delta[n]-C(1.5)^nu[-n-1]+D(0.8)^nu[n]$$
which of course leads to the same result as above. This approach was taken in this answer to your previous question.