When the point of suspension of pendulum is moved, its period of oscillation

When point of suspension of pendulum is moved upwards,

${\mathrm{g}}_{\text{eff }}=\mathrm{g}+\mathrm{a},{\mathrm{g}}_{\text{eff }}>\mathrm{g}\text{ and as }\mathrm{T}\propto 1/\sqrt{{\mathrm{g}}_{\text{eff }}}$, hence T decreases.

i.e., choice (1) is correct.

When point of suspension of pendulum is moved downwards and a > 2g, then the pendulum oscillates with the bob  above the point of suspension and its 

time period becomes $T = 2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{a} - \mathrm{g}}}$. If a $>2g , T decreases.$Hence option (2) is correct .
In case of horizontal acceleration

${\mathrm{g}}_{\text{eff }}=\sqrt{{\mathrm{g}}^2+{\mathrm{a}}^2},\text{ i.e., }{\mathrm{g}}_{\text{eff }}>\mathrm{g}$

i.e., again T decreases.