Nyquists real resistance component
It is a fact that \$R(\omega)\$ depends on \$\omega\$, the proof is as follows:

How do you model a real, physical resistor? Every resistor has a certain, sometimes very small capacitance. How to include it in an equivalent circuit, in parallel or in series? If you put it in series it becomes a highpass, and in reality the noise of any resistor does not show highpass behavior. Thus, it needs to go in parallel, as I drew earlier (see question). 

Now it needs to be put into mathematical equations. In general a parallel circuit is most easily treated with admittances, thus
$$
Y(\omega)=G(w)+jB(w)=G+jB(w).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)
$$
with \$G(\omega)\$ being the inverse of the so called "real resistance component" \$R_0\$. And since \$R_0\$ does not depend on \$\omega\$, \$G(\omega)\$ does not either: \$G(\omega)=G=const\$.

Now Johnson, in historical perspective, preferred to think in impedances. So he converted the admittance into an impedance by taking its inverse or simply considering the parallel circuit of \$R\$ and \$C\$:
$$
Z(\omega)=\frac{1}{Y(\omega)}=R||C=\frac{R\cdot\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{1+j\omega RC}.~~~~~~~~~~~~~~~(5)
$$
Now a little conversion, starting with the multiplication of (5) with \$1=(1-j\omega RC)/(1-j\omega RC)\$:
$$
Z(\omega)=\frac{R\cdot(1-j\omega RC)}{(1+j\omega RC)(1-j\omega RC)}=\frac{R-j\omega R^2C}{1+j(\omega RC)^2}~~~~~~~~~~~~~~~(6),
$$
utilizing \$(a-b)(a+b)=a^2-b^2\$. Now we separate the real part and imaginary part which yields
$$
Z(\omega)=\frac{R}{1+j(\omega RC)^2}-\frac{j\omega RC}{1+j(\omega RC)^2}=R(\omega)+jX(\omega).~~(7)
$$
Thus, the resistive part is frequency dependend, which answers question 1). The answer to question 2) is that the \$R\$ in eq. (7) is really \$R_0\$, so the assumption was correct. The above deduction also answers question 3).