Okay, so now that we've agreed that generalized pseudorandom functions can be of variable length :-D, I'll slightly rephrase your question to meet the definitions I'm working with, and then prove (1).
- Given a pseudorandom function $F = \{F_{n}\}_{n \in \mathbb{N}}$, such that for each $k \in \{0,1\}^n$ , $F_k:\{0,1\}^n \times \{0,1\}^n \to \{0,1\}^n$, is the function defined by $W_{k_1,k_2}(x) = F_{F_{k_1(0 \ ^n )}}(x) || F_{k_2}(x)$ a generalizable pseudorandom function?
Assume by contradiction that $W$ is not a generalizable pseudorandom function,
then there exists a probabilistic polynomial time (PPT) oracle machine $M'$ and a
polynomial $p(\cdot)$ such that for infinitely many $n$'s,
$$\big|\text{Pr}[M'^{W_{k_1,k_2}}(1^{n}) = 1] - \text{Pr}[M'^{H'_{n}}(1^{n}) = 1]\big| > \frac{1}{p(n)}$$
where $k_1,k_2 \xleftarrow{\$} \{0,1\}^n$ and $H' = \{H'_n\}_{n \in \mathbb{N}}$ with $H'_n$ uniformly distributed over the set of all functions mapping $n$-bit strings to $2n$-bit strings.
We will use $M'$ to construct a PPT oracle machine $M$ to distinguish $F_k$, where $k \xleftarrow{\$} \{0,1\}^n$, from $H = \{H_n\}_{n \in \mathbb N}$, where $H_n$ is uniformly distributed over the set of all functions mapping $n$-bit strings to $n$-bit strings. On input $1^n$, $M$ first queries it's oracle $\mathcal{O}$ with the string $0^n$ and records the result as $k' \leftarrow \mathcal{O}(0^n)$. $M$ then uniformly selects $k'' \xleftarrow{\$} \{0,1\}^n$. Next $M$ runs $M'(1^n)$ and serves as its oracle, returning $F_{k'}(x)||F_{k''}(x)$ whenever $M$ queries with input $x \in \{0,1\}^n$. Whatever output $M'$ returns to $M$, $M$ outputs itself.
If $\mathcal{O} = F_k$, then
$$\text{Pr}[M^{F_k}(1^n) = 1] = \text{Pr}[M'^{F_{k'}||F_{k''}}(1^{n}) = 1] = \text{Pr}[M'^{W_{k_1,k_2}}(1^{n}) = 1]$$
since how we chose $k'$ makes our experiment identical from the point of view of $M'$ to when it's interacting with the oracle $W_{k_1,k_2}$. If $\mathcal{O} = H_n$ we have $\text{Pr}[M^{H_n}(1^n) = 1] = \text{Pr}[M'^{F_{H_n(0^n)}||F_{k''}}(1^n) = 1]$. Now, $M'$ is a black box, and was not "designed" to distinguish $W_{k_1,k_2}$ from $F_{H_n(0^n)}||F_{k''}$. That said, we can intuitively see that $F_{H_n(0^n)}||F_{k''}$ should not be distinguishable from $H'_n$ because of our assumption that $F$ is a pseudorandom function. We will now formalize that notion.
Assume that there exists a polynomial $q(\cdot)$ such that for infinitely many $n$'s
$$\big|\text{Pr}[M'^{F_{H_n(0^n)}||F_{k''}}(1^{n}) = 1] - \text{Pr}[M'^{H'_n}(1^{n}) = 1]\big| > \frac{1}{q(n)}$$
We can then use $M'$ to construct a PPT oracle machine $M''$ to distinguish $F_k$ from $H_n$. We denote the oracle of $M''$ by $\mathcal{O}'$. On input $1^n$, $M''$ selects $k''' \xleftarrow{\$} \{0,1\}^n$ and runs $M'(1^n)$, acting as its oracle by returning $\mathcal{O}'(x)||F_{k'''}(x)$ for any query $x \in \{0,1\}^n$ that $M'$ makes.
If $\mathcal{O}' = F_k$, then
$$\text{Pr}[M''^{F_k}(1^n) = 1] = \text{Pr}[M'^{F_k||F_{k'''}}(1^n) = 1] = \text{Pr}[M'^{F_{H_n(0^n)}||F_{k''}}(1^n) = 1]$$
where the last equality follows because $k$ was chosen uniformly at random and $H_n(0^n)$ is a uniformly distributed random variable over the outputs on input $0^n$ of every function mapping $n$ bits to $n$ bits, and $k'''$ and $k''$ were both chosen uniformly at random. In the case $\mathcal{O} = H_n$ we have $\text{Pr}[M''^{H_n}(1^n) = 1] = \text{Pr}[M'^{H_n||F_{k'''}}(1^n) = 1]$. We must now go just one layer deeper before deriving a direct contradiction, and prove by contradiction that $M'$ cannot $H_n||F_{k'''}$ and $H'_n$.
Assume that there exists a polynomial $r(\cdot)$ such that for infinitely many $n$'s
$$\big|\text{Pr}[M'^{H_n||F_{k'''}}(1^{n}) = 1] - \text{Pr}[M'^{H'_n}(1^{n}) = 1]\big| > \frac{1}{r(n)}$$
Since the first $n$ bits of the output of the two oracles should be uniformly distributed, they cannot provide any useful information to $M'$ to make its decision*. Then $M'$ must be able to distinguish $F_{k'''}$ from the second half of $H'_n$ with more than negligible probability. Since this is equivalent to distinguishing $F_k$ from $H_n$, it contradicts our claim that $F$ is a pseudorandom function.
Now we unfold the argument. Since $M'$ cannot distinguish $H_n||F_{k'''}$ and $H'_n$ with more than negligible probability, this contradicts our assumption that $M'$ can distinguish $F_{H_n(0^n)}||F_{k''}$ from $H'_n$ with more than negligible probability. If that assumption is wrong, then $M$ distinguishes $F_k$ from $H_n$ with only negligible difference in probability from that the which $M'$ distinguishes between $W_{k_1,k_2}$ and $H'_n$. By our hypothesis, this is non-negligible, but because we assume $F$ to be a pseudorandom function, no such $M'$ can exist. Therefore, we conclude that $W$ is a generalizable pseudorandom function.
* You could prove this more rigorously by remembering that polynomial-time unpredicability and pseudorandomness imply each other. Then show that you can replace $M'$ with a PPT oracle machine $M'''$ that never reads less than $n+1$ bits of its input (or more than $2n-1$) before guessing a bit, and that the probability that $M'''$ guesses correctly is no less than $M'$.
