What is the impulse response of the system with this difference equation:
Starting from the difference equation
$$y(k)−\frac{5}{12}y(k−1)+\frac{1}{24}y(k−2)=x(k)−\frac{1}{2}x(k−1),$$
you can express the $Z$-transform on both sides ($X(Z)$ and $Y(Z)$ denote the $Z$-transforms of $x$ and $y$) and get:
$$Y(Z)\left(1−\frac{5}{12}Z^{-1}+\frac{1}{24}Z^{-2}\right)=X(Z)\left(1−\frac{1}{2}Z^{-1}\right).$$

Details on this procedure are quite well explained in $Z$ Transform of Difference Equations.

The transfert function $H(Z)$ is given by the ratio of the output by the input:
$$H(Z) = Y(Z)/X(Z) = \frac{1−\frac{1}{2}Z^{-1}}{1−\frac{5}{12}Z^{-1}+\frac{1}{24}Z^{-2}}.$$
As the denominator is (now) a simple polynomial of degree 2 in $Z^{-1}$, you can quite easily expand the rational fraction $H(Z)$ in $Z^{-1}$ to obtain a series $\sum_{k=0}^{+\infty} h(k)Z^{-k}$ by using formal expressions like:
$$\frac{1}{1-X} = 1+X^{-1}+X^{-2}\ldots$$
and
$$\frac{1}{1+aX+bX^2} = c/(X-r_0)+d/(X-r_1)$$
with $r_0$ and $r_1$ the roots of the degree-2 polynomial $1+aX+bX^2$.

With some computations with the numerator, you should be able to retrieve the solution to your expression, with simple formulas for your $h(k)$, providing you with the impulse response you have looking for.

You can find additional material in Partial fraction decomposition. If you prefer a more self-contained, note-like document, I would suggest the following Difference equation solutions from impulse response, which provides an alternative explanation.