= Solution
The <Hahn-Banach theorem> for <bounded linear functionals> says that a <bounded linear functional> on any <linear subspace> of a <normed vector space> extends to the whole space with its <norm> unchanged. No closedness or completeness of the subspace is required. Question 1 uses real-valued $\ell_\infty(\Gamma)$, so its operator assertions are read over the real scalar field. The analogous complex statements use complex-valued indexed functions.
For $x\ne0$, define $g(tx)=t\|x\|$ on its one-dimensional span. This is a norm-one functional, so <Hahn-Banach theorem> supplies an extension $f\in X^*$ with
$$
\boxed{\|f\|=1,\qquad f(x)=\|x\|.}
$$
The <canonical embedding into the bidual> is $Jx(f)=f(x)$. It is linear and $\|Jx\|\le\|x\|$. The <bounded linear functional> just constructed gives the reverse inequality for nonzero $x$, and the zero case is immediate. Thus $\boxed{\|Jx\|=\|x\|}$ and $J$ is injective.
For the <coordinate functional representation of an operator into bounded indexed functions>, let $e_\gamma$ evaluate a coordinate and put $f_\gamma=e_\gamma T$. If $T$ is bounded and linear, then $f_\gamma\in X^*$ and $\|f_\gamma\|\le\|T\|$. Conversely, if $M=\sup_\gamma\|f_\gamma\|<\infty$, the formula $(Tx)(\gamma)=f_\gamma(x)$ defines a bounded scalar function for each $x$, is linear, and satisfies $\|Tx\|_\infty\le M\|x\|$. Combining the two estimates gives
$$
\boxed{T\in\mathcal B(X,\ell_\infty(\Gamma))\iff\sup_\gamma\|f_\gamma\|<\infty,
\qquad\|T\|=\sup_\gamma\|f_\gamma\|.}
$$
For an empty index set both spaces/families have <norm> bound zero; the <supremum> of the empty nonnegative family is taken as zero.
Choose $\Gamma=B_{X^*}$ and $Tx=(f(x))_{f\in B_{X^*}}$. The <Hahn-Banach theorem> makes $\|Tx\|_\infty=\|x\|$, so this is a linear <isometric embedding> into <bounded scalar functions on an index set>.
If $X$ is nonzero and separable, choose a dense sequence $u_n$ in its unit sphere and supporting functionals $f_n$ with $\|f_n\|=f_n(u_n)=1$. For a unit vector $u$ and $\epsilon>0$, some $u_n$ satisfies $\|u-u_n\|<\epsilon$, whence $|f_n(u)|\ge1-\epsilon$. Thus $(f_n)$ is a <countable norming family> and $x\mapsto(f_n(x))$ is an <isometry> into the <l-infinity sequence space>. Completeness of $X$ is not needed.
If instead $X=E^*$ for a separable <normed vector space> $E$, choose a dense sequence $v_n$ in $B_E$. The evaluations $f_n(x)=x(v_n)$ lie in $B_{X^*}$ and continuity of $x\in E^*$ gives $\sup_n|x(v_n)|=\|x\|$. This again gives $\Gamma=\mathbb N$, even when $E^*$ is not separable. If $X=\{0\}$, use zero coordinates throughout.
To prove 1-injectivity, take $T:Y\to\ell_\infty(\Gamma)$ on a subspace of $Z$. Extend every coordinate $f_\gamma\in Y^*$ to $\widetilde f_\gamma\in Z^*$ by <Hahn-Banach theorem>, keeping its <norm>. Their uniform bound defines $\widetilde Tz=(\widetilde f_\gamma(z))$, and the coordinate <norm> identity yields
$$
\boxed{\widetilde T|_Y=T,\qquad\|\widetilde T\|=\|T\|.}
$$
Selection of these extensions uses the usual choice convention for arbitrary index sets. This verifies the <lambda-injective normed space> definition with $\lambda=1$.
For the final <retraction characterization of lambda-injectivity>, suppose first that $X$ is lambda-injective and $i:X\to Z$ is a linear <isometry>. The inverse $i(X)\to X$ has <norm> one when $X\ne0$; extend it to $P:Z\to X$ with $\|P\|\le\lambda$. Then $Pi=I_X$, and $iP$ is a bounded projection onto $i(X)$. For the zero space take $P=0$.
Conversely, suppose every linear <isometry> out of $X$ has such a left inverse. Fix an <isometry> $j:X\to\ell_\infty(\Gamma)$ as above and a left inverse $P$ with $\|P\|\le\lambda$. For any $T:Y\to X$, extend $jT$ to $S:Z\to\ell_\infty(\Gamma)$ by 1-injectivity. Then $\widetilde T=PS$ extends $T$, and $\|\widetilde T\|\le\lambda\|S\|=\lambda\|T\|$. Therefore $X$ is lambda-injective exactly when every isometric embedding $i:X\to Z$ admits a bounded map $P:Z\to X$ satisfying
$$
\boxed{Pi=I_X,\qquad\|P\|\le\lambda.}
$$
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