Complex conjugation of Banach spaces Let $E$ be a Banach space. $\bar{E}$ is the complex conjugation. The following statements are from Pisier's book.
1.$\phi:x\mapsto \bar{x}$ is the identity map from $E$ to $\bar{E}$. Thus, $x$ and $\bar{x}$ are the same element but we "declare " that $\forall \lambda\in \Bbb C$, $\lambda \bar{x}=\overline{\bar{\lambda} x}$.
Does it mean that $\phi(\lambda x)=\bar{\bar{\lambda}}\phi(x)$?
2.For any Hilbert space $H$, $H^*$ can be equipped with the boliorthogonal orthonormal basis. How to interpet this statement.
3.If $a\in B(H, K)$, it is easy to see that $a\mapsto \bar{a}$ is an isomorphism from $\overline{B(H, K)}$ to $B(\bar{H}, \bar{K})$.
We know that $a\in B(H, K)$,   why does $a$ belong to $\overline{B(H, K)}$?
 A: If I understand correctly, the underlying space of $\overline{E}$ is $E$ but scalar multiplication is now multiplication by the complex conjugate. If we denote scalar multiplication on $E$ by concatenation and on $\overline{E}$ by "$\cdot$", the definition of $\overline{E}$ is telling us that $\lambda \cdot x := \bar{\lambda}x$.
However, by some reason (probably to avoid having a special name for scalar multiplication on $\overline{E}$ as I did above) the book you are using has chosen to give a different name to the elements in $\overline{E}$. To be precise, we have $\overline{E}:=\{\bar{x}: x \in E\}$ and scalar multiplication is $\lambda \bar{x}:=\overline{\bar{\lambda}x}$. There an obvious map $\phi: E \to \overline{E}$ which sends $x \mapsto \bar{x}$ and we have 
$$
\phi(\lambda x)= \overline{\lambda x} = \bar{\lambda}\bar{x}=\bar{\lambda}\phi(x)
$$
This tell's us that $E$ and $\overline{E}$ are not quite isomorphic, but conjugate isomorphic (the map $\phi$ is not linear but conjugate-linear). I hope this answers your first question. 
For the second one, I am assuming you meant biorthogonal instead of  boliorthogonal. There is a dual pairing $H \times H^* \to \Bbb{C}$ mapping $(\xi, \omega)\mapsto \omega(\xi)$ for any $\xi \in H$ and $\omega \in H^*$. Some people denote this by $\langle \xi, \omega \rangle := \omega(\xi)$. A biorthogonal orthonormal basis simply means an orthonormal basis $(\xi_j)_{j \in J}$ for $H$ together with $(\omega_{j})_{j \in J}$ such that $\langle \xi_j, \omega_k \rangle = \delta_{j,k}$ for any $j,k \in J$. It's standard Hilbert space theory that you can always find such thing.
Finally, for your third question, as we saw above what we have is that $a \mapsto \bar{a}$ is a conjugate isomorphism from $B(H,K)$ to $\overline{B(H,K)}$. Now there is a map $\Phi:\overline{B(H,K)} \to B(\overline{H}, \overline{K})$ given by $\Phi(\bar{a})(\bar{\xi}):=\overline{a(\xi)}$. This is a composition of two conjugate linear maps and is therefore a linear map. 
