The answer to Q1 is square root of 130.

A = starting vertex

B = farthest point

R = rectangular face including A

Q = rectangular face adjoining R not including A

S = square face not including A

V = vertex not sharing any face with A

W = vertex diagonally opposite V across S

Then arguments by symmetry and contradiction show that B must be on S on the diagonal connecting V and W.

Now the distance from A to B is the smaller of the distances using a route via R then Q then S and using a route via R and S only.

Let the co-ordinates of B on S with respect to origin V be (b,b). (0 <= b <= 4)

Consider the route via RS only and flatten the brick with S adjoining R:

The distance from A to B is the hypotenuse of a right-angled triangle with sides 4-b & 12-b.

Distance via RS = SQRT ((4-b)(4-b) + (12-b)(12-b))

= SQRT (160 - 32*b + 2*b*b)

Differentiating shows this is decreasing in b (in the range 0 <= b <= 4)

Note for b = 0 distance = SQRT(160) and for b = 2, SQRT(104)

Now consider the route via RQS and flatten the brick with S adjoining Q:

The distance from A to B is the hypotenuse of a right-angled triangle with sides 8-b & 8+b.

Distance via RQS = SQRT ((8+b)(8+b) + (8-b)(8-b))

= SQRT (128 + 2*b*b) which is increasing in b

(Actually if b becomes close enough to 4 so that the hypotenuse stops crossing Q then the distance follows a dogleg which is also increasing in b)

Note for b = 0 distance = SQRT(128) and for b = 2, SQRT (136) (since the route is still the hypotenuse at this point)

So the graphs of the 2 distance functions intersect precisely once in [0,2], and this value is the farthest distance.

Solving the the 2 hypotenuse equations simultaneously yields b = 1 and thus AB = SQRT (130) = 11.40175.