The continuity equation is for calculating the change in velocity as your cross sectional area changes.
Bernoulli's equation is for relating various factors of fluid dynamics as a fluid flows through a pipe.
K= P + 1/2pv^2 + pgh
and K in = K out
You solve equations by always assuming that K stays constant. If you flow from point A to point B in a pipe, K at point A must equal K at point B. You can therefore predict qualitative and quantitative change using Bernoulli's equation.
For example. if fluid in a pipe flow horizontally from point A > B and it is more narrow at B, what happens to the pressure? Well first, you can throw out pgh because the height doesn't change from point A to point B. Now, what happens to the velocity when your area decreases?
Use the continuity equation. According to the continuity equation Q=Q, and therefore A1v1=A2v2, and velocity increases with decreasing area. Therefore, velocity at point B is faster than at point A.
Now go back to Bernoulli's equation...and you now know that velocity at B is higher than at A. if K1= K2 , then P + 1/2pv^2 (A) = P + 1/2pv^2 (B). You can now see that if velocity at A is lower than velocity at B, then Pressure at A MUST be higher than Pressure at B. That's a qualitative prediction using Bernoulli's equation - you can obviously plug in numbers to make a quantitative prediction as well.
For your particular question...is it a question involving a large tank of fluid that has a hole that is draining the fluid? That's what it sounds like. The application of Bernoulli's equation isn't as intuitive here, but it works the same way.
At the top of the tank you have: P + 1/2v^2 + pgh (top)
At the bottom of the tank, you have: P + 1/2V^2 + pgh (bottom)
P= pressure pushing on the fluid at that point
v=velocity of the fluid at that point
According to Bernoulli's, you can set these 2 quantities equal to one another:
P + 1/2v^2 + pgh (top) = P + 1/2V^2 + pgh (bottom)
Now, let's make some assumptions about the conditions.
1. Assume that P(top) = P(bottom) - if the container is open to the atmosphere, then P(top)=Patm and P(bottom) = Patm > therefore, you can cancel out the Pressures on both sides of the equation.
2. Assume that the area at the top of the tank is MUCH larger than the area of the hole at the bottom of the tank. A(top) >> A(bottom)
3. Because A(top) >> A(bottom), assume that velocity of the fluid at the top as the hole drains is negligible and v (top) = 0 (think of it like a needle poking a hole in a 2 liter soda bottle...the bottom squirts out fluid, but the fluid height at the top barely seems to change because it's moving SO slowly)
4. Finally, set the level of the hole/drain as zero height (remember, when it comes to potential energy, it's not absolute height that matters, but difference in height)
Now, plug in zero for the P top and bottom, v(top) and h(bottom):
P + 1/2v^2 + pgh (top) = P + 1/2V^2 + pgh (bottom)
pgh (top) = 1/2v^2 (bottom)
solving for v, you will get v=sqrt(2gh)
It seems your problem was a specific application of Bernoulli's equation, one for a tank of fluid draining through a small hole at the bottom. It's not quite the same as a fluid flowing through a pipe because you have to make some assumptions, but it's the same principles.
