Is F_k_1(x) \xor F_k_2(x) a pseudorandom function?
The simplest way to show something like this would be to take advantage of the triangle inequality with a hybrid argument:
$$
\Delta(A, B) \le \Delta(A, C) + \Delta(C, B)\,,
$$
where $C$ is some intermediate construction. 

In this case, you begin with $F_{k_1}(x) \oplus F_{k_2}(x)$, where $k_1$ and $k_2$ are sampled independently and uniformly at random. We can replace $F_{k_1}$ by a truly random function $f_1$, and now we have 
$$
f_1(x) \oplus F_{k_2}\,,
$$
and the distance from the original construction is the distance from $F_{k_1}$ to $f_1$ which is, by definition, $\mathrm{Adv}^{\mathrm{PRF}}_{F_{k_1}}(D_1)$ for some distinguisher $D_1$ that makes $q$ queries to $F_{k_1}$ and has runtime $t$.

Doing the same for $F_{k_2}$, we have
$$
f_1(x) \oplus f_2(x)\,
$$
adding the distance $\mathrm{Adv}^{\mathrm{PRF}}_{F_{k_2}}(D_2)$ similarly as before. Since the xor of two uniformly random strings is just as uniform as a single one, we can move to the random function
$$
f_3(x)\,,
$$
without any degradation in advantage. The total advantage is the sum of all these steps:
$$
\mathrm{Adv}^{\mathrm{PRF}}_{F_{k_1,k_2}}(D) \le \mathrm{Adv}^{\mathrm{PRF}}_{F_{k_1}}(D_1) + \mathrm{Adv}^{\mathrm{PRF}}_{F_{k_2}}(D_2) + 0\,.
$$