For establishing the inequality $e^x>1+x$, consider two cases: $x>0$ and $x<0$.
For the case $x>0$, apply the Mean Value Theorem to the function $f(x)=e^x$ over the interval $[0,x]$. This gives a $c$ with $0<c<x$ satisfying
e^x-e^0 =(x-0)\cdot e^c=xe^c.
But $e^0=1$ and $e^c>1$ (strict inequality, since $c>0$). so
which implies the result for $x>0$.
I'll leave the other case for you...
I'll just give a hint for your second inequality:
For the second inequality, break things up into two cases: $x\ge 0$ and $x<0$.
For the case $x\ge0$, apply the Mean Value Theorem to the function $g(x)=2x\arctan x -\ln(x^2+1)$ over the interval $[0,x]$ and use the fact that $\arctan(c)\ge 0$ for all $c\ge0$. (Note: the second case is easier here, since $g$ is an even function.)