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 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. 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).