(a) No. One of the two colors, say green, will occupy at least $\lceil (3\cdot 7)/2\rceil = 11$ squares. Let $g_i$, $1\leq i \leq 7$ be the number of green squares on column $i$. Then the total number of pairs of green squares on a same column, over all $7$ columns, is $\sum_{i=1}^7 \binom {g_i} {2} = \sum_{i=1}^7 \frac {g_i(g_i - 1)} {2}$. The function $f(x) = \dfrac {x(x-1)} {2}$ is convex, so $\sum_{i=1}^7 f(g_i) \geq 7f\left (\dfrac {\sum\limits_{i=1}^7 g_i} {7}\right ) \geq 7f\left (\dfrac {11} {7}\right ) = \frac {22} {7} > 3$, hence at least $4$. But there are only $\binom {3} {2} = 3$ possibilities, so two columns will share a pair of green squares positioned on same rows, thus creating a subrectangle with green corner squares. (b) Yes. $\begin{array} {|c|c|c|c|c|c|} \cline{1-6} & & * & & * & * \\ \cline{1-6} & * & & * & & * \\ \cline{1-6} * & & & * & * & \\ \cline{1-6} * & * & * & & & \\ \cline{1-6} \end{array}$