I answer Question 1. It is just a calculation.

Instead of a real torus, say ${\bf T}$, I consider a pair $(T,\sigma)$,
where $T$ is a complex torus and $\sigma\colon T\to T$ is an anti-holomorphic involution.
See [this question](https://mathoverflow.net/q/342300/4149) and YCor's answer.

For a complex torus $T$, consider the *cocharacter group*
 $${\sf X}_*(T)={\rm Hom}(T, {\Bbb G}_{m,{\Bbb C}}).$$
To a real torus ${\bf T}=(T,\sigma)$ we associate a pair ${\sf X}_*({\bf T}):=({\sf X}_*(T),\sigma_*)$,
where $\sigma_*\in {\rm Aut\,}\,{\sf X}_*(T)$ is the induced automorphism.
It satisfies  $\sigma_*^2=1$.

We denote $\Gamma={\rm Gal}({\Bbb C}/{\Bbb R})=\{1,\gamma\}$, where $\gamma$ is the complex conjugation.
We obtain an action of $\Gamma$ on ${\sf X}_*(T)$ (namely, $\gamma$ acts via $\sigma_*$).
In this way we obtain an equivalence between the  category of ${\Bbb R}$-tori
and the category of $\Gamma$-lattices (finitely generated ${\Bbb Z}$-free $\Gamma$-modules):
$$ {\bf T}\rightsquigarrow {\sf X}_*({\bf T}). $$
Moreover, this is an exact functor:  a short exact sequence of real tori
$$ 1\to{\bf T}'\to{\bf T}\to{\bf T}''\to 1$$
induces a short exact sequence of $\Gamma$-lattices
$$ 0\to {\sf X}_*({\bf T}') \to {\sf X}_*({\bf T}) \to {\sf X}_*({\bf T}'')\to 0.$$

Now consider the torus ${\Bbb G}_{m,{\Bbb R}}=({\Bbb C}^\times,\,z\mapsto\bar z)$ and the corresponding $\Gamma$-lattice  $({\Bbb Z},1)$.
Moreover, consider the torus 
$$R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}=(\,{\Bbb C}^{\times\,2},\, (z_1,z_2)\mapsto (\bar z_2,\bar z_1)\,)$$
and the corresponding $\Gamma$-lattice $({\Bbb Z}^2,J)$,
where 
$$ J=\begin{pmatrix}0&1\\1&0\end{pmatrix}. $$
Consider the norm homomorphism
$$N\colon R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}\to {\Bbb G}_{m,{\Bbb R}},\quad (z_1,z_2)\mapsto z_1z_2$$
and the corresponding morphism of $\Gamma$-lattices
$$N_*\colon ({\Bbb Z}^2,J)\to ({\Bbb Z},1),\quad (x_1,x_2)\mapsto x_1+x_2.$$
By definition, 
$$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}=\ker N,$$
and so its cocharacter group is $\ker N_*=\{(x, -x)\mid x\in{\Bbb Z}\}.$
The complex conjugation $\gamma$ acts on $\ker N_*$ by $J$, that is, 
$$(x,-x)\mapsto (-x, x).$$
We see that $\ker N_*\simeq ({\Bbb Z},-1)$, and hence 
$$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}\simeq ({\Bbb C}^\times, z\mapsto \bar z^{\,{-1}}).$$
Since
$$ (z\mapsto \bar z^{\,{-1}})\,=\,(z\mapsto z^{-1})\,\circ\,(z\mapsto \bar z),$$
we see that $R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}$ can be obtained from ${\Bbb G}_{m,{\Bbb R}}=({\Bbb C}^\times,\,z\mapsto\bar z)$
by twisting by the cocycle $\gamma\mapsto (z\mapsto z^{-1})$, as required.

Note that these three $\Gamma$-lattices 
$({\Bbb Z},1),\ ({\Bbb Z}^2,J),$, and $({\Bbb Z},-1)$  are the only indecomposable $\Gamma$-lattices (up to isomorphism); 
see [this answer](https://mathoverflow.net/a/27145/4149). 
It follows that these three real tori ${\Bbb G}_{m,{\Bbb R}}$, 
$R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}$, and 
$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}$
are the only indecomposable real tori (again, up to isomorphism).