This was intended as a comment, but for the sake of clarity, I'd better use an answer.
Regarding to the case $\mu \neq 1$, we can start using the following set of equations, which are derived from the Maxwell equations and after applying boundary conditions that demand that across the boundary the tangential components of $E$ and $H$ should be continuous.
$$\cos\theta_{i}(A_{\parallel}-R_{\parallel})=\cos\theta_{t}T_{\parallel}$$
$$A_{\perp}+R_{\perp}=T_{\perp}$$
$$\sqrt{\frac{\epsilon_{1}}{\mu_{1}}}\cos\theta_{i}(A_{\perp}-R_{\perp})=\sqrt{\frac{\epsilon_{2}}{\mu_{2}}}\cos\theta_{t}T_{\perp}$$
$$\sqrt{\frac{\epsilon_{1}}{\mu_{1}}}(A_{\parallel}+R_{\parallel}=\sqrt{\frac{\epsilon_{2}}{\mu_{2}}}T_{\parallel}$$
Then, adding together the first and fourth equation, you obtain
$$T_{\parallel}=\frac{2\cos\theta_{i}\sqrt{\epsilon_{1}\mu_{2}}}{\cos\theta_{t}\sqrt{\mu_{2}\epsilon_{1}}+\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}}A_{\parallel}$$
Adding the second and third equation, you have
$$T_{\perp}=\frac{2 \sqrt{\mu_{2}\epsilon_{1}}\cos\theta_{i}}{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}+\cos\theta_{i}\sqrt{\mu_{2}\epsilon_{1}}}A_{\perp}$$
Accordingly for $R_{\parallel}$ and $R_{\perp}$ (in which we have to substitute the value we already found for $T_{\parallel}$ and $T_{\perp}$)
$$R_{\parallel}=\frac{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}-\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{t}}{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}+\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{t}}A_{\parallel}$$
$$R_{\perp}=\frac{\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{i}-\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}}{\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{i}+\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}}A_{\perp}$$