TITLE: Three old questions on the Sacks forcing
QUESTION [18 upvotes]: I came across the two following Qs in 1970.

Find reals $a,b$ such that $a$ is Sacks over $L[b]$ and vice versa $b$ is Sacks over $L[a]$. Note that a Sacks $\times$ Sacks generic pair definitely does not work.
Suppose that $a$ is Cohen over $L$ and $b$ is Sacks over $L[a]$. Does $a$ belong to $L[b]$? Comment: this Q answers in the positive if $a$ is Sacks or Solovay-random over $L$. 

The next Q is perhaps not that dead lock, but still I have no clue.

Let $\mathbf  P$ consist of all perfect sets $P$ on the real plane $\mathbb R\times\mathbb R$ such that all vertical and all horisontal cross-sections of $P$ are perfect (in particular, non-countable). What does it force?

REPLY [9 votes]: For (2), I suggest writing $a$ as the Turing join of its even and odd parts, call them $a_0$ and $a_1$.  Use $a_0$ to compute a perfect subtree $T$ of $2^{<\omega}$ such that every infinite path in $T$ is mutually Cohen generic with $a_1$.  When considered as a Sacks condition in $L[a]$, $T$ forces that $a$ is not an element of $L[b]$.