Solution:

1. Consider the upper half plane, that is the set \[ V = \left\{ (x,y)\in \mathbb{R} ^2 : x\geq 0 \right\}. \] Note that the set $V$ is closed under addition but not under scalar multiplication. For that, if $\left( x_1,y_1 \right), \left( x_2,y_2 \right) \in V $, then \begin{align*} \left( x_1 ,y_1 \right) + \left( x_2, y_2 \right) & = \left( x_1+x_2, y_1+y_2 \right). \end{align*} Since $x_1 \geq 0$ and $x_2\geq 0$, so the sum $x_1+x_2 \geq 0$ and hence it is closed under addition. On the other hand, we have $(1,1)\in V$ but $-1\cdot (1,1) = (-1,-1)\notin V$. Therefore, $V$ is not closed under scalar multiplication.

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2. For a subset which is closed under scalar multiplication but not under addition we consider the following set \[ W = \left\{ (x,x):x\in \mathbb{R} \right\} \cup \left\{ (x,2x): x\in \mathbb{R} \right\} \eqqcolon W_1 \cup W_2. \] To show it is closed under scalar multiplication we take an element from $W$, so it will be either from $W_1$ or $W_2$. In any case, we note that for any $\alpha \in \mathbb{R} $, \begin{align*} \alpha \cdot (x,x) = (\alpha x,\alpha x) \in W \\ \alpha \cdot (x,2x) = (\alpha x,2\alpha x) \in W . \end{align*} Therefore, it is closed under scalar multiplication. Now we will show that it is not closed under addition. For that we take $(1,1), (1,2) \in W$. Note that \[ (1,1) + (1,2) = (1,3) \notin W. \] Therefore, $W$ is not closed under addition.