$(a)$ $\because$ The rods have identical dimensions.

Let their area of crossection be $=A$ and length be $=L$

So each rod would have heat resistance of

$R=\frac{L}{K A}$

$R_{\text {eff }}=R_1+R_2+R_3$

$=\frac{L}{K_{A l} A}+\frac{L}{K_{C u} A}+\frac{L}{K_{A l} A}$

$R_{\text {eff }}=\frac{L}{A} \times \frac{5}{400}$   $\left[\because K_N=200, K_{C u}=400\right]$

$H_1=\frac{\Delta T}{R_{\text {eff }}}=\frac{100}{\frac{L}{A} \times \frac{5}{400}} \quad \ldots (1)$

$(b)$ When rods are connected in parallel

$\frac{1}{R_{\text {eff }}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$R_{\text {eff }}=\frac{L}{A} \times 800$

$H_2=\frac{100}{\frac{L}{A} \times 800} \quad \ldots (2)$

$(2)$ by $(1)$

$\frac{40}{H_2}=\frac{400}{800 \times 5} \Rightarrow H_2=400\,W$