Conservation of Mechanical Energy

At the highest point the swing momentarily stops; at the lowest point the speed is maximum.

Apply conservation of mechanical energy between the maximum and minimum points:
$mgh = mgh_1 + \dfrac{1}{2}mV^2$
$V = \sqrt{2g(h - h_1)} = \sqrt{2 \times 10 \times (2 - 0.75)} = \sqrt{25}$
$V = 5$ m/s

On a frictionless incline only gravity does work (the normal force does no work).
By conservation of energy, for both the man and the bag: $mgh = \dfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$
The speed depends only on the height fallen, not on the path or the mass. (The times taken are different.)
So $V_B = V_m$.

At angle $\theta$, the bob is at height $h = l - l\cos\theta = l(1 - \cos\theta)$ above the lowest point.

Potential energy at the extreme position = kinetic energy at the mean position:
$mgl(1 - \cos\theta) = \dfrac{1}{2}mv^2$
$v = \sqrt{2gl(1 - \cos\theta)}$