Finding the radius of a sphere touching other spheres inscribed in n-dimensional spaces I know for certain that the radius of the first one is $r=(\sqrt2-1)/2$. I assume the radius of the other dimensions are the same but I don't know how I would create an equation to prove that. Lastly I have no idea what the last part is asking, I would assume it would never leave the cube but I don't understand how higher dimensions look like so I cannot confirm that either.
"In the $2\times2$ square inscribe $4$ equal disks of radius $\frac12$. Then in the center inscribe one more disk touching all $4$ disks. Compute its radius.
Similar in $3D$ space: in $2\times2\times2$ cube inscribe $8$ balls of radius $\frac12$ and in the center inscribe a ball touching all $8$ balls in the corners. Compute its radius.
Similarly, go to $4D$, $5D$, $nD$ space.
Problem: find smallest $d$ for which the central $d$-dimensional ball touching $2n$ $\frac12$-radius balls will stick out of $2\times2\times...\times2$ $n$-dimensional cube"
 A: This question seems complicated given that it's hard to visualize spaces with more dimensions than $\mathbb{R}^3$ has, but the method to find the radius of the central ball is the same for each. The calculation doesn't get any trickier as we go up in dimension, and I'll walk you through how to do it.
Step 1 - place your $n$-dimensional box such that one corner is on $(0, ...,0)$, and the opposite one is on $(2,...,2)$. Imagine the segment $D$ connecting these points, and use the distance formula in $\mathbb{R}^n$ to express its length as a function of $n$.
Step 2 - we expect the central circle to have points of tangency with two circles on the line segment $D$, and it passes through the diameters of the two circles. If we call the radius of the central circle $r$, then we know so far that $r = \frac12$(|D| - 2 - 2A), where the 2 is the length of D in each circle of radius $\frac12$, and $A$ is the length from a corner to a circle. We can calculate $A$ by finding the distance from the center of a circle to the nearest corner, and subtracting $\frac12$ (the radius). We only need the distance formula in $\mathbb{R}^n$ here; and I recommend expressing $A$ as a function of $n$ as well.
Step 3 - Plug your formulas for $D$ and $A$ into $\frac12(|D| - 2 - 2A)$, and you'll have the general radius of the ball in $\mathbb{R}^n$.
Step 4 - PROFIT
It helps to draw a picture of the 2-dimensional case; as I said before the argument in that dimension is exactly the same in each higher dimension.
