Solutions to Piet Hein's puzzles
No matter what Red does, Blue can connect via either g5 or g8. Both blue stones at the right are connected to the right (see the Strategy guide for details). The stone at d8 is connected to the left because of the ladder breaker at c2.
Using twice edge template III 2b
Blue 1 is connected to the right via Template IIIa and threatens to connect directly to the left. The only possibility to prevent this connection is to play out the ladder:
Since the cells for the ladder and those for the edge template do not overlap, Red cannot do anything against the connection.
We interpret "optimal play" to mean that the winning player tries to win in as few moves as possible, and the losing player tries to postpone the loss as long as possible.
The unique opening move for which the board will be completely filled is Red a2 (or equivalently on the other side of the board, Red c2). It leads to the following sequence of forced moves:
At this point, Red has already won (due to the double threat at b2 and c1). To postpone the loss as long as possible, Blue should play c1, which forces Red b2, followed by 4 more moves to fill in the edge templates.
No other opening move fills the board completely. If Red opens at b2, they win in at most 4 more moves. If Red opens at a1 or b1, Blue plays b2 and wins in at most 4 more moves. If Red opens at c1, optimal play proceeds as follows:
This ends with a win for Red in at most 2 more moves. The remaining opening moves are symmetric to the ones already discussed.
The two blue pieces are connected to the right using edge template III2a. Red must block the connection to the left at B2
And now blue can connect with either E2 or E1. Had red played 4. E3 blue could respond with D4 with a similar outcome. Had red played 2. D3 to block the future ladder, blue could respond with 3. C4, keeping the connection to the right and connecting to the left in two different ways (via B2 and B5).