Note I don't know how the artificial gravity is produced, but I am assuming you can selectively shape a potential gravity well so it affects only the things in the ship; the payload. There are problems with this kind of artificial gravity, and to make this answerable I am going to ignore them. Additionally, the individual forces exerted on the payload will be in various directions for various situations, and I assume the craft is capable of this.
Coasting
It could just create a gravity well, with the Energy around that of Earth. The equation to figure this out would look like:
$$U=\frac{-Gm_E}{r_E}\sum{m_n}$$
where $G$ is the universal gravitational constant, $m_E$ is the mass of the earth, and $r_E$ is the radius of the earth, $m_n$ is the mass of an object you want to feel this force, and U is the energy of the gravity well. In your specific example $\sum{m_n} = 1000$. (The ship itself does not need to be under the effect of this gravity well, so it doesn't get added to $\sum{m_n}$ at all.) So your gravity well looks like it needs to be $6.249*10^{10}$(-ish) joules deep.
The craft itself contributes to the gravity well, but only a little. The more mass of the craft is centered on one side, the more it can can offset the gravity needed to keep people down. I have ignored the craft's gravity because it's negligible influence. If the ship were a ball that people stood on, with their center of masses 1m away, it would only contribute about $.003 J$. Not enough to substantially affect the gravity well.
Trying to determine the power requirements for bending space enough or producing this gravity well is tricky, because no one has every produced a gravity well like this before, nor have we seen gravity (of an appreciable amount) apply to bodies and then not. This is where sci-fi magic/handwaving comes in. To simply provide an answer, I am going to say that you need $6.249*10^{10} W$: you need to maintain that gravity well by giving it the required energy for the total depth every second.
Obviously, this value can change depending on your views of how artificial gravity production works, but it's the answer I'm running with. If you think that, once a gravity well is made, you needn't give it more energy to maintain it, your power requirements go to 0 after it's made. If you think gravity wells must apply their energy over plank time, then you'll get a very large power requirement (Around $32*10^{54}$ Watts!)
Maneuvering
For the people inside the ship to notice no acceleration, the ship's inertial dampener must compensate for the movement of the ship. The well will have to get deeper or shallower by the amount of work needed to keep the people on the floor. This depends greatly on which way the ship accelerates. The equation for this work, though presents a problem:
$$W=\int_{x_1}^{x_2}{Fdx}$$
No, the problem is not the calculus, it is the fact the work required depends on how long the acceleration is experienced. If you attempt to solve this for our specific situation, you get $$W=\int_{x_1}^{x_2}{5dx}=5*(x_2-x_1)=5*\Delta x$$
That's not hard. Your well changes by 5 times the distance (in meters) traveled. That being said, you don't need to produce this energy all at once; your power requirements may not change all that much. The total fuel you need and the total energy you need to produce will!
The power demand with fluctuate depending on how much you travel per second. It would look like: $$P = \frac{5*\Delta dx}{dt} = -5*v$$
That $v$ is velocity. This changes, of course, with how fast you're going. Since you're not jumping from one speed to the next every second, the above equation gives you the instantaneous power you need. I suggest you look at your top speed, as that is the maximum amount of power you need to reach that top speed.
Under Attack
1 Gigajoule $(10^9 J)$ just got added to the mix! I need to assume that this energy is radiated in a sphere, not just directly impacting the ship. This makes the impact profile of the ship really important. The surface area of a sphere of radius $10m$ is about $1256.4 m^2$ Our serendipitously cylindrical space ship has a side surface area of $10m*\pi*3m=94.24...m^2$. This means the ship gets a dose of ($94.24/1256.4=.075...$) about 8% of the gigajoule, or $8*10^{7} J$. Once again, the gravity well needs to increase by this amount to prevent our payload from getting harmed.
The power needed for this? Well, it depends on how long the explosion lasts. If the explosion lasts a small fraction of a second, you multiply that $8*10^{7}$ by the reciprocal of that fraction. An explosion which transfers its energy over 1/100 of a second, for example, will require $8*10^{9}$ Watts to totally negate.