a)Choose $1,2,4,5$ to get $k=4$ b) We will prove, that $k=4$ is maximal. Let $k \geq 5$ then pair $a+b, a+c$ or $b+d,c+d$ should be in AP $a+b<a+c<b+c<b+d<c+d$ and $a+b<a+c<a+d<b+d<c+d$ Let $a+b, a+c$ are first two elements of AP, then $b+c$ and $a+d$ can not third, because it leads to $a+c=2b$ or $b+d=2c$ . So In such case not more than $4$ is possible Same logic is for last two elements $c+d, b+d$ So answer is $4$

@RagvaloD for b) 1 2 3 form an arithmetic progression. (The actual answer to b) is $4$.)

@above Thanks, I fixed