The correct thermodynamic conditions for the spontaneous reaction at all temperatures is

Option-1 is correct:

$\mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}$

$\begin{array}{l}\text{ If }\mathrm{\Delta H}<0\text{ and }\mathrm{\Delta S}>0\\ \mathrm{\Delta G}=(-\mathrm{ve})-\mathrm{T}(+\mathrm{ve})\end{array}$

$\mathrm{then} \mathrm{at} \mathrm{all} \mathrm{temperatures}, \mathrm{\Delta G}=-\mathrm{ve}, \mathrm{spontaneous} \mathrm{reaction}.$

Also, option-3 also correct:

$\text{ If }\Delta H<0\text{ and }\Delta S=0$

$\Delta G=(-\mathrm{ve})-T\left(0\right)=-\mathrm{ve}\text{ at all temperatures. }$

Basically, this question has two correct options-1 and 3. Also, according to answer key published by NTA , 1 and 3 are correct answers.