If a focal chord to $y^2=16x$ is tangent to $(x-6{)}^2+y^2=2$, then the possible value(s) of the slope of this chord is/are

$\because$ focal chord of the parabola $y^2=16x$ will always passes through focus i.e., $(4,0)$, which lies outside to the given circle
So, the question can be termed as slope of tangents drawn from $(4,0)$
Now the circle is $(x-6{)}^2+y^2=(\sqrt{2}{)}^2$
$\therefore r=\sqrt{2}$

From the diagram, we have
$\sin \theta =\frac{r}{2}=\frac{1}{\sqrt{2}}\Rightarrow \theta ={45}^{\circ }$
Therefore, slope of the chords are $=\pm \tan 45°=\pm 1$.

Alternate Solution:
Equation of any focal chord of the parabola $y^2=16x$ is
$y=m(x-4)$
Now for the above line to be tangent of the circle $(x-6{)}^2+y^2=2$
distance of the line from center should be equal to radius
$\frac{\left|2m\right|}{\sqrt{1+m^2}}=\sqrt{2}\Rightarrow 2m^2=(1+m^2)\Rightarrow m^2=1$
Hence slope $=m=\pm 1$