By
vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by

where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by

where

is its direction with respect to the x-axis.
Hi there!
We can use the rotational equivalent of Newton's Second Law:

Στ = Net Torque (Nm)
I = Moment of inertia (kgm²)
α = Angular acceleration (rad/sec²)
We can plug in the given values to solve.

Answer:
a=2500J,b=1000K,c=1000J,d=14.142m/s
Explanation:
V²=U²+2gh
V²=0 + 2×10×10=200m/s
a).kinetic energy=(1/2)mv²=(1/2)25×200=2500
potential energy=mgh
p.e=25×10×10=2500J
pe+ke=2500+2500=5KJ
b).mgh=25×10×4=1000J
c). V²=U²+2gh
V²=0+2×10×4
V²=80
kinetic energy=(1/2)mv²
=(1/2)25×80
=1KJ
d). From my first paragraph V²=200
V=√200
V=14.142m/s
Personally, I agree with your answer, namely that the likely-intended event happening here is one of acceleration. Having said that, I also want to add: it pains me to see this type of wording because, clearly, it is vague and only invites confusion of the type you are talking about.
Good luck!