Given: $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$

Parabola can be written 

${\left(y-2\right)}^2=4\left(x-1\right)$

Equation of tangent in slope form 

$\left(y-2\right)=m\left(x-1\right)+\frac{a}{m}$

$\Rightarrow y=mx-m+\frac{1}{m}+2 ...\left(1\right)$  $\left(a=1\right)$

Now given tangent $y=\frac{x}{2}+\frac{k}{2} ...\left(2\right)$

Using Equations $\left(1\right) \& \left(2\right)$

$m=\frac{1}{2} \& \frac{k}{2}=2-m+\frac{1}{m}$

So $k=7$

Now slope at $\left(1, 7\right)$

$\frac{dy}{dx}=\frac{2}{y-2}=\frac{{\displaystyle 2}}{{\displaystyle 7-2}}=\frac{2}{5}$