Let the resistances of the two conductors be $R_1$ and $R_2$ respectively. Given that the resistances are connected in series, their equivalent resistance is: $$ \begin{aligned} R_1 + R_2 &= 40 \quad \dots (i) \end{aligned} $$ When connected in parallel, the equivalent resistance is $7.5 \, \Omega$. Thus: $$ \begin{aligned} \frac{1}{R_1} + \frac{1}{R_2} &= \frac{1}{7.5} \\ \Rightarrow \frac{R_1 + R_2}{R_1 R_2} &= \frac{1}{7.5} \end{aligned} $$ Substituting the value of $R_1 + R_2$ from equation $(i)$: $$ \begin{aligned} \frac{40}{R_1 R_2} &= \frac{1}{7.5} \\ \Rightarrow R_1 R_2 &= 40 \times 7.5 \\ \Rightarrow R_1 R_2 &= 300 \quad \dots (ii) \end{aligned} $$ From equation $(i)$, we have $R_2 = 40 - R_1$. Substituting this into equation $(ii)$: $$ \begin{aligned} R_1(40 - R_1) &= 300 \\ \Rightarrow 40R_1 - R_1^2 &= 300 \\ \Rightarrow R_1^2 - 40R_1 + 300 &= 0 \\ \Rightarrow (R_1 - 30)(R_1 - 10) &= 0 \end{aligned} $$ Therefore, $R_1 = 30$ or $R_1 = 10$. If $R_1 = 30$, then $R_2 = 10$. If $R_1 = 10$, then $R_2 = 30$. Thus, the resistances of the two conductors are $30 \, \Omega$ and $10 \, \Omega$.