Do I understand impedance correctly? $$Let\quad \vec { Z } =R+Xj$$
If $X>0$,then the impedance is lagging (current lags behind voltage).If $X<0$, then the impedance is leading (current leads voltage or more accurately, voltage lags behind current). If $X=0$, the current and voltage are in phase and the load is called purely real. If $X\neq 0$, the current and voltage are not in phase and the load is called complex.
Some load is purely real if and only if the load is purely resistive.
Some load is purely imaginary if and only if the load is purely reactive.
Questions:


*

*Does it make physical sense for a load to be purely imaginary or purely reactive?

*How do we know that when $X>0$, current lags behind voltage and not the other way around?

*Is everything I have said correct?
 A: 
Does it make physical sense for a load to be purely imaginary or
  purely reactive?

An ideal resistor is a purely real load.  An ideal inductor or capacitor is a purely imaginary load.
Any physical resistor, inductor or capacitor will possess resistance, inductance, and capacitance and thus, a self-resonance frequency.
For example, at frequencies "high enough", we must model a physical resistor as, e.g.

However, for low frequency applications, the reactive part of a resistor's impedance is insignificant and, to the precision one typically works to, is a pure resistance.

How do we know that when X>0, current lags behind voltage and not the
  other way around?

Let the phasor voltage $V_s$ be the voltage across some impedance $Z = R + jX = |Z|e^{j\phi}$ where $|Z| = \sqrt{R^2 + X^2}$ and $\tan \phi = \frac{X}{R}$ and $X > 0$
The resulting phasor current is
$$I_s = \frac{V_s}{Z} = \frac{V_s}{|Z|}e^{-j\phi}$$
See that, due to the negative sign in the exponent, the current phasor lags the voltage phasor (the negative sign means rotate the current phasor clockwise by $\phi$ relative to the voltage phasor).
