Solution: Given that $f:\mathbb{R}\to \mathbb{R} $ satisfies $f(x+y)=f(x)+f(y)$ for each $x,y\in \mathbb{R} $.

1. Note that for any $x\in \mathbb{R} $ and $n\in \mathbb{N} $ we have \begin{align*} f(nx) & = f(x+ x+\cdots+x) \\ & = \underbrace{f(x) + f(x) + \cdots + f(x)}_{n\text{-times}} \\ & = n f(x). \end{align*}

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2. We need to show that $f$ is continuous at a single point if and only if $f$ is continuous on $\mathbb{R}$. If $f$ is continuous on $\mathbb{R} $, then it will be continuous at a single point. So we only need to show that if $f$ is continuous at a single point then it is continuous on $\mathbb{R} $. Let us assume that $f$ is continuous at $a \in \mathbb{R}$ and $x_0\in \mathbb{R} $ be any point. We claim that $f$ is continuous at $x_0$.
   Note that \begin{align*} \lim_{x \to x_0} f(x) & = \lim_{x \to a} f \left( x -a +x_0 \right) \\ & = \lim_{x \to a} \left[ f\left( x+\left( x_0-a \right) \right) \right] \\ & = \lim_{x \to a} \left( f(x) + f\left( x_0-a \right) \right) \\ & = \lim_{x \to a} f(x) + \lim_{x \to a} f\left( x_0-a \right) \\ & = f(a) + f\left( x_0 - a \right) \\ & = f \left( a + x_0 -a \right) \\ & = f \left( x_0 \right). \end{align*} Thus, $f$ is continuous at $x_0$.

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3. Note that if $f(x)=\alpha x$ for some $\alpha \in \mathbb{R} $, then it is continuous. Now we will prove that if $f$ is continuous then $f(x)=\alpha x$ for some $\alpha \in \mathbb{R} $. Observe that \[ f(0) = f(0+0) = f(0) + f(0) \implies f(0) = 0. \] Similarly, \[ f(0) = f(x-x) = f(x) + f(-x) \implies -f(x) = f(-x). \] So for any $n\in \mathbb{N} $, we have \begin{align*} f(n) = f(1 + \cdots + 1) = n f(1) \\ f(-n) = -f(n) = -n f(1). \end{align*} Let us set $f(1) = \alpha $. Then we have for any $n\in \mathbb{Z} $ \[ f(n) = \alpha n. \] Now we take $x\in \mathbb{Q} $, so $x=\frac{p}{q}$ for some $p,q\in \mathbb{Z} $ and $q\neq 0$. Note that, \begin{align*} q\cdot f\left( \frac{1}{q} \right) = f\left( q\cdot \frac{1}{q} \right) = f(1) \implies f\left( \frac{1}{q} \right) = \frac{\alpha}{q}. \end{align*} So we have \begin{align*} f(x) & = f\left( \frac{p}{q} \right) \\ & = f \left( p\cdot \frac{1}{q} \right) \\ & = p f\left( \frac{1}{q} \right) \\ & = \frac{p}{q}\cdot \alpha \\ & = \alpha x. \end{align*} Therefore, we proved that for any rational number $x$ $f(x) = \alpha x$.
   
   Now let $x$ be an irrational number. Since, rationals are dense in $\mathbb{R} $, we can find a sequence $\left( x_n \right) \subseteq \mathbb{Q} $ such that $x_n\to x$. As $f$ is continuous, $f(x_n)$ must converge to $f(x)$. Note that \[ f\left( x_n \right) = \alpha x_n \to \alpha x. \] Therefore, \[ f(x) = \alpha x. \] Hence, we proved that if $f$ is continuous, then $f(x)=\alpha x$ where $\alpha =f(1)$.