There are no polarity problems, and indeed a current has no direction, because it's a scalar quantity. What makes circuit theory complicated with the signs of the various involved quantities is that they are integral laws, and that's, why there are various "directions" involved, which lead to the signs of these various quantities.
For AC circuit theory involving only "passive elements" like resistors, capacitors, and coils, you just deal with Faraday's Law in integral form taking the line integral along each loop of your circuit and charge conservation at each node.
I'm not sure, which circuit you are really discussing here. From #1 it seems to be a capacitor and a coil in series connected to an AC voltage ##V(t)=V_0 \cos(\omega t)##.
In the above diagram the arrow indicates the orientation of the line integral in Faraday's Law,
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
This line integral gives, taking into account the chosen "polarities" of the voltage source and the capacitor,
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-V(t)+C Q.$$
The magnetic flux has to be calculated with the surface-normal vector going into the plane and in this direction also is the magnetic field. So you have
$$-\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}=-L \dot{i}.$$
Together with ##i=\dot{Q}## you get
$$-V(t)+C Q=-L \ddot{Q}$$
or
$$L \ddot{Q} + C Q=V(t)=V_0 \cos(\omega t).$$
It's of course a bit artificial, assuming that there's no resistance (at least the coil in reality has some resistance, which you can put in series with the ideal coil which adds a term ##R i=R \dot{Q}## on the left-hand side of the equation).