You can do problem 1) in Cartesian coordinates or in spherical coordinates. Let me give an outline of how you can do it in Cartesian coordinates first.

The potential energy is:

U = gamma/|r1 - r2|

We will calculate minus the gradient and then compare that with the force law to see if minus the gradient is indeed the force. In Cartesian coordinates you can write:

|r1 - r2| = sqrt[(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]

The components of the gradient are the derivatives with respect to the coordinates. You can take the gradient w.r.t. to r1 = (x1,y1,z1) or r2 = (x2,y2,z2). If we calculate the gradient w.r.t. r1 and take the x-component then we have to differentiate U w.r.t. x1. Let's first rewrite U as:

U = gamma [(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]^(-1/2)

dU/dx1 = -1/2gamma[(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]^(-3/2)*2(x1-x2) =

-gamma[(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]^(-3/2)*(x1-x2).

The term in the square brackets is |r1-r2|^2, so yo can write this in a "mixed" form as:

dU/x1 = -gamma/|r1-r2|^3 (x1-x2)

Similarly you find:

dU/y1 = -gamma/|r1-r2|^3 (y1-y2)

dU/z1 = -gamma/|r1-r2|^3 (z1-z2)

These are the three components of the gradient. You can reconstitute any vector from it's components by multiplying each component by the appropriate unit vector and adding those terms up. If V is a vector and the components are Vx, Vy and Vz, then:

V = Vx x-hat + Vy y-hat + Vz z-hat.

So, we have:

Nabla1 U = dU/x1 x-hat + dU/y1 y-hat + dU/z1 z-hat =

-gamma/|r1-r2|^3 [ (x1-x2)x-hat + (y1-y2)y-hat + (z1-z2)z-hat ]

The term in the square brackets is r1 - r2. We can write:

[r1 - r2]/|r1-r2|^3 = e_{1,2}/|r1-r2|^2

where e_{1,2} is the unit vector that points in the direction of r1 from r2.

This means that:

-nabla1 U = gamma e_{1,2}/|r1-r2|^2

In the gravitational case gamma = -MmG and in the electrostatic case gamma = kq1q2, and you can see that you get the correct force law in these cases.

If you calculate the gradient w.r.t. r2, then the only difference will be a minus sign because U is a function of (r1 - r2) and the chain rule will yield a minus sign. This means that:

-nabla2 U = -gamma e_{1,2}/|r1-r2|^2

but -e_{1,2} = e_{2,1} which is the unit vector that points in the direction of r2 from r1. So, you again obtain the correct force law.

You can also derive this result in spherical coordinates. If you take the point r2 as the origin then:

U = gamma/r

What are the components of the gradient in spherical coordinates? For a general function
F(r, theta, phi) we have:

dF = dF/dr dr + dF/dtheta dtheta + dF/d phi dphi (1)

But dF must also be given by:

dF = nabla F dot ds

where ds is the displacement vector. ds can be written in terms of its components in spherical coordinates as:

ds = dr r-hat + r dtheta theta-hat + rsin(theta) dphi phi-hat

Note that r dtheta is the distance in the theta hat direction, because if you change theta, you are rotating in that direction with a radius of r. If you change phi, you are rotating about the z-axis with a radius of r sin(theta) therefore the distance in the phi-hat direction is r sin(theta) dphi.

The inner product nabla F dot ds is the r-hat component of nabla F times the r-hat component of ds plus the theta hat component of nabla F times that component of ds plus the phi component of nabla F times that component of ds. But the result must be formula (1). From that you deduce that you must have:

nabla F = dF/dr r-hat + 1/rdF/dtheta theta-hat + 1/(r sin(theta))dF/dphi phi-hat

In our case U does not depend on theta or phi, so, we only have an r-hat component which is given by the derivative w.r.t. r:

nabla U = dU/dr r-hat U = -gamma r-hat/r^2 --->

-nabla U = gamma r-hat/r^2

which is the same result we found above.

Problem 2:

U(r1,r2,r3,r3) = U12(r1-r2) + U13(r1-r3)+U14(r1-r4)+U23(r2-r3)+U24(r2-r4)+U34(r3-r4)+U1(r1)+U2(r2)+U3(r3)+U4(r4).

We have:

-dU/dr3 = -d/dr3 U13(r1-r3) - d/dr3 U23(r2-r3) - d/dr3 U34(r3-r4) - d/dr3 U3(r3) =

We have seen in problem 1) that the above derivatives are the forces exerted on particle 3 in case of the inverse square law force. In general, minus the derivative of the potential energy is the force.