Edit: Sorry, I made a stupid mistake in the previous post, here is the (hopefully) correct solution.

Hmm... I'm not very good at explaining things but this one is pretty easy. All you have to know is:
1) which forces affect the stone (one - gravity)
2) what are the variables (s - vertical coordinate, v - velocity, g - acceleration) and their signs (s and v are positive (aimed upwards), g is negative (aimed downwards))
2) how the stone moves (as Newton determined, it has a constant acceleration g ~= 10 (m/s/s))
3) if its starting velocity is v0, what its velocity is at a moment of time t1 (v1 = v0 - g*t1, by definition of movement with constant acceleration)
4) if its starting velocity is v0 and its starting vertical coodinate is s0, what its vertical coordinate is at a moment of time t1 (s1 = s0 + v0*t1 - g*t1*t1/2, again by definition)

When the stone hits the ground, its vertical coordinate s1 is zero. We have a quadratic equation, solving it we find out t1 (when it will be on the ground). And since the stone started moving at t0 = 0, t1 will also be the total time in the air.

Knowing that, it's trivial to find the terminal velocity v1.

As for the formulas you've given, they might be true or they might not. It depends on how you define the symbols (what is u? what is X?) and how you apply them to the problem (why do you think "s = ut + 0.5 X at^2" or "v^2 = u^2 + 2as" would give you something?). If in doubt, it's always better to use the well-known laws than trying to figure if a particular formula works in your case.

In: v0 = 25 (m/s), s0 = 2 (m), g = 10 (m/s/s); s(t1) = 0, t1 >= 0;
Out: t1? v(t1)?

From condition (b) s(t1) = s0 + v0*t1 - g*t1*t1/2 = 0.
Solve the equation: t1 = (v0+-sqrt(v0*v0+2*g*s0))/g = (25+-sqrt(625+40))/10 ~= 5.078 (dropping the negative root).

v1 = v0 - g*t1 = 25 - 10*5.078 ~= 2.031