Because $f\in H^1(0,T;H^{-1})$, hence $f\in C(0,T;H^{-1})$, you may infer that $t\mapsto f(t)$ is continuous into $L^2$ equipped with its weak topology. To prove that it is continous into $L^2$ equipped with its strong (normed) topology, you need that $t\mapsto\|f(t)\|$ be continous ; this is not guaranted by your assumptions, essentially because the spaces $H^{1/2}$ and $H^{-1}$ are not in duality.

Because $f\in H^1(0,T;H^{-1})$, hence $f\in C(0,T;H^{-1})$, you may infer that $t\mapsto f(t)$ is continuous into $L^2$ equipped with its weak topology. To prove that it is continous into $L^2$ equipped with its strong (normed) topology, you need that $t\mapsto\|f(t)\|$ be continous ; this is not guaranted by your assumptions, essentially because the spaces $H^{1/2}$ and $H^{-1}$ are not in duality.

Edits. Because we already know that $t\mapsto f(t)$ is continuous from $(0,T)$ into $L^2_w$, we see that the continuity into $L^2$ for the strong topology is equivalent to the continuity of $t\mapsto\|f(t)\|_L^2$. This is why $f\in L^\infty(0,T;H^{-1/2})$ doesn't help you.