(a)(i) Wheatstone bridge balance condition: at balance, no current flows through the galvanometer, so equal currents $I_1$ flow through arms $P$, $Q$ and $I_2$ through arms $R$, $S$; with zero potential difference across the galvanometer:

$$I_1P=I_2R,\quad I_1Q=I_2S \Rightarrow \frac{P}{Q}=\frac{R}{S}$$

(ii) This requires the specific resistor/emf values in the given circuit figure, which were not fully legible in the extracted question. Apply the balance condition and Kirchhoff's rules to the actual circuit to find the current in the $3\ \Omega$ branch.

(b)(i) From $I=nAev_d$ and $v_d=\dfrac{eE\tau}{m}$: current density $J=\dfrac{I}{A}=nev_d=\dfrac{ne^2\tau}{m}E$. Comparing with Ohm's law $J=\sigma E$:

$$\sigma=\frac{ne^2\tau}{m}$$

(ii) $R_1=1.05\ \Omega$ at $20°C$, $R_2=1.38\ \Omega$ at $100°C$

$$R_2=R_1[1+\alpha(100-20)]$$

$$1.38=1.05[1+80\alpha] \Rightarrow \frac{1.38}{1.05}=1.314=1+80\alpha$$

$$\alpha=\frac{0.314}{80}\approx3.93\times10^{-3}\ °C^{-1}$$