The questions asks to demonstrate that the solution to the difference equation $(y_{n+1} - 2 y_{n} + y_{n-1}) - \frac{1}{2} h (y_{n+1} - y_{n-1}) - 2 h^2 y_{n} = 0$ with initial condition $y_0 = 1$ with $y_n$ bounded as $n \rightarrow \infty$ is $y_n = (1 - h + \frac{1}{2} h^2)^n$.
Following the normal procedure of deriving a characteristic equation and solving for the roots results in the messy expression $ \frac{2+2h^2 \pm h^2 \sqrt{9+4h^2}}{2-h}$. How can this solution be simplified? Moreover, how does guaranteeing than $y_n$ be bounded as $n \rightarrow \infty$ change the solution?
The equation is a discretization of $$ 0=y''-y'-2y=(D+1)(D-2)y $$ so that the general solution is $y(x)=Ae^{2x}+Be^{-x}$. Thus it is expected that the roots you get are $1+2h+O(h^2)$ and $1-h+O(h^2)$. As that is not the case, your root formula is wrong, with a high likelihood you made an error by extracting the common power of $h$ from the square root.
You will want the coefficient for the unbounded basis solution to be zero, it remains the basis solution for the root $$ \frac{2(1+h^2)-h\sqrt{9+4h^2}}{2-h} = \frac{2(1+h^2)-3h(1+\frac29h^2-\frac2{81}h^4+O(h^6))}{2-h} \\ =1-h+\frac{h^2+O(h^3)}{2-h}=1-h+\frac{h^2}2+O(h^3) $$