EXTENDED/REVISED ANSWER
Some general points concerning the second question. By definition, $m$ is a median of $X$ if ${\rm P}(X \ge m) \geq 1/2$ and ${\rm P}(X \le m) \geq 1/2$. While a median is uniquely determined for any common example of a continuous random variable, it is not uniquely determined in general. For example, any number $m \in [-1,1]$ is a median for random variable $X$ with ${\rm P}(X=1) = {\rm P}(X=-1) = 1/2$. Hence my previous answer to this question (see below), where I assumed that $m(X)=0$ since $X$ is symmetric, should be revised. This is done simply as follows. We define $X$ and $Y$ exactly as before, and introduce another random variable $\tilde X$ defined to be equal to $X$ with probability $1-1/n$ and to $0$ with probability $1/n$. It is immediately checked that the symmetric random variables $\tilde X$ and $Y$ have a unique median, equal to $0$. Thus $m(\tilde X) + m(Y) = 0$, as required. On the other hand, one easily verifies that ${\rm P}(\tilde X + Y = 1) \to 3/4$ as $n \to \infty$ (cf. my previous answer), which implies that $\tilde X + Y$ has a unique median, equal to $1$. So, $m(\tilde X + Y) \neq m(\tilde X) + m(Y)$, as required.
In view of this example, we now give a counterexample for the case where $X$ and $Y$ are independent. Let $X$ and $Y$ be i.i.d. random variables with common probability mass function given by $p(2)= p(-1) = \frac{1}{2}(1 - \frac{1}{n})$, $p(0) = \frac{1}{n}$. Then, $X$ and $Y$ have a unique median, equal to $0$. On the other hand, one verifies that both ${\rm P}(X+Y \geq 1)$ and ${\rm P}(X+Y \leq 1)$ tend to $3/4$ as $n \to \infty$; hence, $X + Y$ has a unique median, equal to $1$. So, $m(X + Y) \neq m(X) + m(Y)$, as required.
In view of the preceding examples, we finally consider the case where $X$ and $Y$ are both symmetric and independent. Assuming both $X$ and $Y$ have a unique median, it must obviously be equal to $0$. For any fixed numbers $a$ and $b$, $aX + bY$ is also symmetric. Moreover, $aX + bY$ has a unique median, equal to $0$. This can be carried out straightforwardly, upon observing that ${\rm P}(X \in (-\varepsilon,\varepsilon), Y \in (-\varepsilon,\varepsilon)) > 0$ for any $\varepsilon > 0$. Hence, $m(aX+bY)=am(X)+bm(Y)=0$. From this, it easy to establish the following generalization. Suppose that $X$ and $Y$ are independent and symmetric around arbitrary points, say $m_1$ and $m_2$, respectively. Assume that both $X$ and $Y$ have a unique median (these medians are necessarily given by $m(X)=m_1$ and $m(Y)=m_2$). Then, for any fixed numbers $a$ and $b$, $m(aX+bY)$ has a unique median, equal to $am(X)+bm(Y)=am_1+bm_2$.
PREVIOUS ANSWER
For the second question, let us show that even if $X$ and $Y$ are symmetric random variables, then $m(X+Y)$ might be different from $m(X)+m(Y)$ (where $m$ denotes median). Suppose that ${\rm P}(X=1) = {\rm P}(X=-1) = 1/2$; hence $X$ is symmetric. Define $Y$ as follows: if $X=1$ then $Y=0$, whereas if $X=-1$ then $Y=2$ or $-2$ with probability $1/2$ each. Then, ${\rm P}(Y=0)=1/2$, ${\rm P}(Y=2)=1/4$, and ${\rm P}(Y=-2)=1/4$. Hence $Y$ is symmetric and, in turn, $m(X)+m(Y)=0+0=0$. However, ${\rm P}(X+Y=1) = 3/4$ (and ${\rm P}(X+Y=-3) = 1/4)$. In particular, $m(X+Y)=1$ (since ${\rm P}(X+Y=1) \geq 1/2$).