Solution 1:

The convex hull of four points consists of either $3$ or $4$ points, and this is also the cardinality of the largest subset of points forming a convex polygon. If the probability of $4$ points to form a convex quadrilateral is $p$, the expected cardinality of the largest subset is $p\cdot4+(1-p)\cdot3=3+p$.

The probability that $4$ points independently uniformly distributed in a square form a convex quadrilateral is given in the MathWorld article on Sylvester's four-point problem (along with various generalizations) as $25/36$. Thus the expected cardinality of the largest subset is $133/36$.