Electrical – In a transformer, what is the phase difference b/w secondary induced voltage and current in secondary circuit for purely resistive load

What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -

Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -

\$V_P\cdot I_P = V_S\cdot I_S\$

The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.

This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.

The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.

To answer your questions: We don't choose anything with transformers to have positive polarity. See the wiki. You could reverse the dots on both sides and still get the same result. The dots help you to know what the physical relation of the transformer coils are. If you have two coils, you have four terminals to connect to the circuit. But the circuit doesn't care which side is positive or negative. It only cares if the coils are reversed or not. It is just a notation. Notations are only useful if you know the rules behind them and how they are realized in the model and the physical world.

As far as magnetic flux goes, you really need to think about the magnetic fluxes in terms of mutual inductance. The transformer model uses 4 inductances, 2 of which are mutual (usuually called M) and characterize the flux between the two inductors. Vp always exists across the primary, you have to have a voltage to have current, it may be zero if its a time varying field.

Maintaining high flux in the core applies to non-ideal transformers. An ideal transformer has no resistance in the coil and no magnetic resistance in the material of the core. The idea there is if you introduce a material you have losses, you also have losses to outside world. In a universe with only a transformer, all the field lines would loop back into the other end, and you would have no losses. In the real world they connect and do work on other materials. In addition, air is not a great way to conduct magnetic fields, it has a low magnetic permeability. So we use a good material like iron to conduct the most of the magnetic field with its high magnetic permeability. This also has its problems because iron saturates, it has a point where it can't conduct very well. This saturation must be taken into account or it will clip the top of the sine wave.

I think a better way to look at a transformer is you are trading off voltage for current. A power engineer might be concerned about the inner workings of a transformer where you have losses (and they actually have magnetic circuits with multiple coils) but in the ideal case all you have to worry about are the mutual inductance, the inductance of the coils themselves and the ratio of the coils.

And the last paragraph doesn't make much sense to me either. So I can't answer your question there.