For an example of $2$, you took the usual metric on $d_2$ and made is so that a sequence $x_n$ which is going to $\infty$ becomes Cauchy by shrinking the distances. This is exactly the right idea. The function $f:(-1,1)\to \mathbb{R}$ given by $$f(x)=\frac{x}{1-|x|}$$ also works, and has inverse $$f^{-1}(x)=\frac{x}{1+|x|}.$$ This is just your function composed with a map to take $(0,1)$ to $(-1,1)$.
For the other case, assume $d_1(x,y)\leqslant d_2(x,y)$ for all $x,y\in M$. Assume that $(M,d_1)$ is complete and $(M,d_1)$ is not. Then if $(x_n)_{n=1}^\infty$ is $d_2$-Cauchy, this means $$\lim_N\sup_{m,n\geqslant N}d_1(x_m,x_n)\leqslant \lim_N\sup_{m,n\geqslant N}d_2(x_m,x_n)=0,$$ because $(x_n)_{n=1}^\infty$ is $d_2$-Cauchy. Therefore it is also $d_1$-Cauchy, and therefore $d_1$ convergent to some $x$. What we would like is that $(x_n)_{n=1}^\infty$ is not convergent to $x$ in the $d_2$ metric, so $d_2(x_n,x)$ stays large.
So what's to stop us from making all the other points of $M$ far away from $x$, simply by defining $d_2$ to make it so?
More specifically, take $M=[0,1]$. Let $d_1(x,y)=|x-y|$. Define $d_2(x,y)=d_1(x,y)$ whenever $x,y\neq 0$. Of course, $d_2(0,0)=0$. We need to define $d_2(0,x)=d_2(x,0)$ for $x\in (0,1]$ in order to make, for example, $d(0,1/n)$ uniformly large. How can we define $d_2(0,x)$, $x\in (0,1]$, in order to make the resulting $d_2$ actually be a metric (it'll need to be positive definite, so we don't want to do something silly like make $d_2(0,x)=0$) and it'll need to satisfy the triangle inequality (so $d_2(x,z)\leqslant d_2(x,y)+d_2(y,z)$ whenever $0\in\{x,y\}$). And the values $d(0,x)$ should not converge to zero as $x\to 0^+$. Can we do that?
