If $\Omega$ is bounded, then yes. If $\Omega$ is unbounded, then no. If $\Omega$ is bounded and $\varphi\in C^0(\bar{\Omega})$, then $\varphi$ is bounded on $\bar{\Omega}$ and hence $\varphi\in L^2(\Omega)$.

If $\Omega$ is bounded, then yes. If $\Omega$ is unbounded, then no. If $\Omega$ is bounded and $\varphi\in C^0(\bar{\Omega})$, then $\varphi$ is bounded on $\bar{\Omega}$ and hence $\varphi\in L^2(\Omega)$. If $\Omega$ is any unbounded domain, then you can find even a $C^\infty$ smooth radial function $\varphi(x)=f(|x|)$ that growths so fast that $\varphi\not\in L^2(\Omega)$.

If $\Omega$ is bounded, then yes. If $\Omega$ is unbounded, then no.

If $\Omega$ is bounded, then yes. If $\Omega$ is unbounded, then no. If $\Omega$ is bounded and $\varphi\in C^0(\bar{\Omega})$, then $\varphi$ is bounded on $\bar{\Omega}$ and hence $\varphi\in L^2(\Omega)$.