Here is a complete proof of 36
Problem 36 - Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_n\}_{1}^{\infty}$ is a countable dense subset of the unit ball of $\mathcal{X}$, and define $T:L^{1}(\mu)\rightarrow \mathcal{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_n$.
a.) $T$ is bounded.
b.) $T$ is surjective.
c.) $\mathcal{X}$ is isomorphic to a quotient space of $L^{1}(\mu)$ (Use Exercise 35).
Proof a.) - We have that $Tf = \sum_{1}^{\infty}f(n)x_n$, hence $\| Tf\| = \| \sum_{1}^{\infty}f(n)x_n\| \leq \sum_{1}^{\infty}|f(n)|\| x_n\| \leq \sum_{1}^{\infty}|f(n)| = \| f\|$ Thus $T$ is bounded.
Proof b.) - Since $T$ is linear, it's enough to show that every $x\in X$ with $||x||\leq 1$ is in the image of $T$. To do this, we proceed inductively: there exists $x_{n_1}$ such that $||x-x_{n_1}||<\frac{1}{2}$. If $y=2(x-x_{n_1})$ then $||y||<1$, so there exists $x_{n_2}\neq x_{n_1}$ such that $||y-x_{n_2}||<\frac{1}{2}$, hence $||x-x_{n_1}-\frac{1}{2}x_{n_2}||<\frac{1}{4}$. In general, if $x_{n_1},\dots,x_{n_k}$ have been chosen such that $ \Big|\Big|x-\sum_{j=1}^k2^{1-j}x_{n_j}\Big|\Big|<2^{-k}$ then $y=2^k(x-\sum_{j=1}^k2^{1-j}x_{n_j})$ is in the unit ball, hence there exists $x_{n_{k+1}}\not\in\{x_{n_1},\dots,x_{n_k}\}$ such that $||y-x_{n_{k+1}}||<\frac{1}{2}$, i.e. $ \Big|\Big|x-\sum_{j=1}^{k+1}2^{1-j}x_{n_j}\Big|\Big|<2^{-k-1}$ Finally, define $f\in \ell^1(\mathbb{N})$ by $f(n)=2^{1-k}$ if $n=n_k$, $f(n)=0$ otherwise. Then $Tf=x$.
proof c.) - We have that $L^{1}(\mu)$ and $\mathcal{X}$ are Banach spaces and $T\in L(L^{1}(\mu),\mathcal{X})$. Moreover, $T$ is surjective, so $\textrm{range}(T)= \mathcal{X}$ and $\mathcal{X}$ is closed in $\mathcal{X}$. So by exercise 35. $L^{1}(\mu)/\mathcal{N}(T)$ is isomorphic to $\mathcal{X}$.