Problem 1388

Call the $8$ unit segments along the outside of the figure outer edges, and call the remaining $4$ unit segments inner edges. The count can be organized by the number of red inner edges.
* If all $4$ of the inner edges are red, then in order for each square to be bounded by $2$ red segments and $2$ blue segments, all the outer edges must be blue. This gives $1$ coloring.

* If $3$ of the inner edges are red, then there are $4$ ways to choose which inner edge is blue, there is only $1$ way to color the outer edges of the $2$ squares not bounded by the blue inner edge, and there are $2$ ways to color the outer edges of each of the $2$ squares that are bounded by the blue inner edge. This gives $4 \cdot 2 \cdot 2 = 16$ colorings.

* If $2$ of the inner edges are red and these edges form a $90^\circ$ angle (as in the figure), then there are $4$ ways to choose these $2$ red inner edges, $2$ ways to color the outer edges of $2$ of the squares, but only $1$ way to color the outer edges of the other $2$ squares. Again this gives $4 \cdot 2 \cdot 2 = 16$ colorings.

* If $2$ of the inner edges are red and these edges are collinear, then there are $2$ ways to choose these $2$ red inner edges and $2$ ways to color the outer edges of each of the $4$ squares. This gives $2 \cdot 2^4 = 32$ colorings.

By symmetry, the number of colorings with $1$ or $0$ red inner edges is the same as the number of colorings with $3$ or $4$ red inner edges, respectively. The final count is therefore $1 + 16 + 16 + 32 + 16 + 1 = 82$.