e) Let $f: \{0\} \to \mathbb{R}^n$ and $g:\mathbb{R}^n \to \{0\}$. Then $g \circ f = id_{\{0\}}$ since $f$ is the inclusion map. But $f \circ g=f(x)\ \forall x \in \mathbb{R}^n$. How do I show $g$ is homotopic to $id_{\mathbb{R}^n}$?
f) Any hints on how to start here?
For (e), if I understand correctly, you have defined $f:\{0\}\to\mathbb R^n$ by $f(0)=0$, and $g:\mathbb R^n\to\{0\}$ by $g(x)=0$ (the only possible function). Thus, $g\circ f(0)=g(0)=0$, which is the identity function on $\{0\}$. Furthermore, $f\circ g(x)=f(0)=0$. We need to construct a homotopy $H:\mathbb R^n\times I\to\mathbb R^n$ such that $H(x,0)=0$ and $H(x,1)=x$. For this, we might try the straight-line homotopy $H(x,t)=tx$. Can you see why this works?
For (f), note that there are only two possible functions from $\{0\}$ to $\{0,1\}$. Call them $f_0$ and $f_1$, where $f_0(0)=0$ and $f_1(0)=1$. Let $g:\{0,1\}\to \{0\}$ be given by $g(x)=0$. Clearly whether $i=0$ or $i=1$, $g\circ f_i(0)=g(i)=0$, which is the identity function on $\{0\}$. But $f_1\circ g(x)=f_1(0)=1$. We need a homotopy $H:\{0,1\}\times I\to\{0,1\}$ such that $H(x,0)=1$ and $H(x,1)=x$. In particular, $H(0,t)$ is a continuous function from $1$ to $0$. Can you prove that this means that $H(0,t)$ must be discontinuous? The argument for $f_0$ is symmetric.