In the system shown in the figure the mass m moves in a circular arc of angular amplitude 60°. Mass 4m is stationary. Then

Tension in the string will be maximum when mass m is at point B, so

$\mathrm{T}=\mathrm{mg}+\frac{{\mathrm{mv}}^2}{\mathrm{\ell }}$

By Law of Conservation of Energy,

$\begin{array}{l}\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}=\mathrm{mg\ell }(1-\cos \mathrm{\theta })\\ \Rightarrow {\mathrm{v}}^2=2\mathrm{g\ell }\left(1-\cos \left({60}^{\circ }\right)\right)\\ \Rightarrow \mathrm{T}=\mathrm{mg}+\frac{2\mathrm{mg\ell }}{\mathrm{\ell }}\left(1-\frac{1}{2}\right)\\ \Rightarrow \mathrm{T}=2\mathrm{mg}\end{array}$

For mass 4m to be stationary, $\mathrm{f}\ge \mathrm{T}$

$\begin{array}{l}\Rightarrow \mathrm{\mu }\left(4\mathrm{mg}\right)\ge 2\mathrm{mg}\\ \Rightarrow \mathrm{\mu }\ge 0.5\end{array}$

Hence, (1) is correct
When m moves from A to B it has vertical displacement downwards, i.e., along the direction of gravitational force
So, (2) is correct.
Tension is always perpendicular to the velocity, so power delivered by the tension is zero. 

So, (3) is correct.
According to Work Energy Theorem,
${\mathrm{W}}_{\text{total }}=\mathrm{\Delta K}$
Since work done by tension is zero therefore work done only by the gravitational force. So, (4) is also correct.