Don't understand

(a)

$\sqrt{x^2+y^2}\ =\ 5$

The solutions to this equation are all points with a distance of  5  from the origin.

So this is the equation of a circle with a radius of 5 centered at the origin.

I used a little help from here: https://math.stackexchange.com/questions/518856/integral-points-on-a-circle

There is a Pythaogrean triple with a hypotenuse of 5

A triangle with side lengths 3, 4 and 5 is a Pythagorean triple. So...

From the origin, we can go over 3 units and either up or down 4 units to reach an integer solution.

From the origin, we can go over 4 units and either up or down 3 units to reach an integer solution.

From the origin, we can go over 0 units and either up or down 5 units to reach an integer solution.

From the origin, we can go over 5 units and either up or down 0 units to reach an integer solution.

Here's a graph showing all integer solutions:   https://www.desmos.com/calculator/txwzj5nmt4

There are  12  integer solutions.

b? Do I need to make a 3d graph?

(b) How many ordered triples (x,y,z) of integers are there such that \sqrt{x^2 + y^2 + z^2} = 7? Does the question have a geometric interpretation?

I  get the following  triples

6, 3, 2    

But each of these can take on pos/ neg   values for each integer   so there are 2^3  = 8 possibilities  and each of these can be arranged in 3! ways  = 6 ways....so  8 * 6  = 48  possibilities

And

7 0  0       the "7" can be pos/negative   = 2 possibilities  and can be in any 3 positions  = 2 * 3  = 6 possibilities

So.....(If I haven't missed any)  we have 48 + 6   =  54  points [ just as you found ]

This is a sphere centered at  (0, 0, 0)   with a radius of 7

Here's a (not so good) image :