Step 1
Given vector:\[w = \left\{ {\left[ {\begin{array}{*{20}{l}}x\\y\end{array}} \right]:xy \ge 0} \right\}\]

Step 2: (a)
We are given that if ${\bf{u}}$ is in V, then we have to prove that $c{\bf{u}}$ is in V for all scalers.

If, \[{\bf{u}} = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\]then the product:\[\begin{array}{l}c{\bf{u}} = c\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\\\\c{\bf{u}} = \left[ {\begin{array}{*{20}{c}}{cx}\\{cy}\end{array}} \right]\end{array}\]Since ${\bf{u}}$ is in V, we have:\[xy \ge 0\]so for any scaler c:\[\begin{array}{l}xy \ge 0\\\\\left( x \right)\left( y \right) \ge 0\\\\{c^2}\left( x \right)\left( y \right) \ge 0\\\\\left( {cx} \right)\left( {cy} \right) \ge 0\end{array}\]Hence, the vector $c{\bf{u}}$ is in V.

Step 3: (b)
let us take a set of vectors that lie in V\[\begin{array}{l}{\bf{u}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 24}\end{array}} \right]\,::\left\{ { - 3 \times - 24 = 72 > 0} \right\}\\\\{\bf{v}} = \left[ {\begin{array}{*{20}{c}}{10}\\1\end{array}} \right]\,\,::\left\{ {10 \times 1 = 10 > 0} \right\}\end{array}\]Addition of the vectors\[\begin{array}{l}{\bf{u}} + {\bf{v}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 24}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{10}\\1\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{ - 3 + 10}\\{ - 24 + 1}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}7\\{ - 23}\end{array}} \right]\end{array}\]To Check if the above vector lies in V, we add its elements
\[7 - 23 = - 16 < 0\]Hence, the vector ${\bf{u}} + {\bf{v}}$ does not lie in V