Here is a solution to the first problem using four pieces. Piece d is the 4×4 square and the 3×3 square is divided into pieces a, b, and c as shown.

Here is a solution to the second problem using eight pieces that is due to R.F. Wheeler. It appeared in Eureka in 1950. Piece a is the 3×3×3 cube. Pieces b and c make up the 4×4×4 cube. The rest of the pieces form the 5×5×5 cube as shown in the second diagram below.

