Let $E\subseteq\mathbb{R}$ be measurable with $\mu(E)<\infty$ and $$d(f, g)=\int_E\dfrac{|f-g|}{1+|f-g|}dx$$ for all measurable $f, g:E\to\mathbb{R}$.
For all measurable $f, g, h:E\to\mathbb{R}$, $d(f, g)\ge 0$ with equality if and only if $f=g$ almost everywhere in $E$, $d(f, g)=d(g, f)$, and $d(f, h)+d(h, g)\ge d(f, g)$.
Let $\{f_k\}$ be a sequence of measurable functions from $E$ to $\mathbb{R}$. $\lim_{k\to\infty}d(f_k, f)=0$ if and only if $f_k$ converges in measure to $f$ on $E$.
1 was straightforward. $\tfrac{|f-g|}{1+|f-g|}\ge 0$ for all $x$, so $d(f, g)\ge 0$. $d(f, g)=0$ if and only if $\tfrac{|f-g|}{1+|f-g|}=0$ almost everywhere if and only if $f=g$ almost everywhere. $d(f,g)=\int_E\tfrac{|f-g|}{1+|f-g|}dx=\int_E\tfrac{|g-f|}{1+|g-f|}dx=d(g, f)$. The triangle inequality reduces to $|f-g|+|g-h|+\text{nonnegative terms}\ge |f-h|$.
I don't really know how to approach 2. I expect showing that $d$ being a metric is relevant, but I can't see how. I was thinking $\lim_{k\to\infty}d(f_k, f)=0$ implies $\lim_{k\to\infty}\int_E|f_k-f|dx=0$ since the latter is at most the former for all $k$, and you might be able to prove the reverse. But apart from that, I'm stuck.
Suppose that $f_k$ converges in measure to $f$. Use
$$\frac{|x|}{1+|x|} \leq \min\{1,|x|\}$$
to show that
$$d(f_k,f) \leq \int_{|f_k-f| > \epsilon} \, \mu(dx) + \int_{|f_k-f| \leq \epsilon} \epsilon \, \mu(dx) \leq \mu(|f_k-f|> \epsilon) + \epsilon \mu(E).$$
Deduce that $d(f_k,f) \to 0$ as $k \to \infty$.
Suppose that $d(f_k,f) \to 0$ as $k \to \infty$. Use that for fixed $\epsilon>0$
$$m_{\epsilon} := \inf_{|x| \geq \epsilon} \frac{|x|}{1+|x|} >0$$
to show that
$$\begin{align*} \mu(\{x \in E; |f_k(x)-f(x)| > \epsilon\}) &= \frac{1}{m_{\epsilon}} \int_{|f_k-f|>\epsilon} m_{\epsilon} \, \mu(dx) \\ &\leq \frac{1}{m_{\epsilon}} \int_{|f_k-f|>\epsilon} \frac{|f_k(x)-f(x)|}{1+|f_k(x)-f(x)|} \, \mu(dx) \\ &\leq \frac{1}{m_{\epsilon}} d(f_k,f). \end{align*}$$