= Solution
Fix $z_0\notin\Gamma$ and put $\delta=\operatorname{dist}(z_0,\Gamma)>0$. If $|z-z_0|<\delta/2$, then
$$
\left|\frac1{w-z}-\frac1{w-z_0}\right|
\leq \frac{2|z-z_0|}{\delta^2}
$$
uniformly for $w\in\Gamma$. Multiplication by the bounded <continuous function> $f$ and integration therefore prove continuity of $F$ at $z_0$.
Take $\Gamma$ to be the positively oriented unit circle and $f\equiv1$. By the <Cauchy integral formula>, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.
For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. <Integrals> over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours $\Gamma_j\subset\Omega\setminus K$, and their sum is
$$
f(z)=\sum_j\frac1{2\pi i}\oint_{\Gamma_j}\frac{f(w)}{w-z}\,dw
$$
for every $z\in K$.
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