The plate current in a triode valve is observed to be zero, even...

The plate current in a triode valve is determined by the effective potential $V_{eff} = V_p + \mu V_g$, where $V_p$ is the plate voltage, $V_g$ is the grid voltage, and $\mu$ is the amplification factor.

The plate current $I_p$ is zero when the effective potential is less than or equal to zero, i.e., $V_p + \mu V_g \le 0$.

Let us analyze the given options:

For $V_g < 0$ and $V_p > 0$, $I_p = 0$ is possible if the grid is sufficiently negative such that $\mu |V_g| \ge V_p$ (cut-off condition).

For $V_g > 0$ and $V_p < 0$, $I_p = 0$ is possible if the plate is sufficiently negative such that $|V_p| \ge \mu V_g$.

For $V_g < 0$ and $V_p < 0$, both the grid and the plate repel the electrons, so $V_p + \mu V_g < 0$ is always satisfied, and $I_p = 0$.

For $V_g > 0$ and $V_p > 0$, the effective potential is always positive, so $I_p$ cannot be zero.

Therefore, the situations where the plate current can be zero are given by options (1), (3), and (4).