Solution:

Initial Value,

$ \begin{aligned} x(0) & =\operatorname{lt}_{z \rightarrow \infty} x(z) \\\\ & =\operatorname{lt}_{z \rightarrow \infty} \frac{1+z^{-1}}{1-0.25 z^{-2}}\\ \end{aligned}\\ $

$ =\frac{1+\infty^{-1}}{1-0.250^{-2}}=\frac{1+0}{1-0}=1\\ $

$x(0)=1$

Final Value,

$ \begin{aligned} x(\infty) & =\operatorname{lt}_{z \rightarrow 1}\left(1-z^{-1}\right) \times(z) \\\\ & =\operatorname{lt}_{z \rightarrow 1}\left(1-z^{-1}\right) \frac{\left(1+z^{-1}\right)}{1-0.25 z^{-2}}\\ \end{aligned} $

$ =\operatorname{lt}_{z \rightarrow 1} \frac{1-z^{-2}}{1-0.25 z^{-2}}\\ $ …