Let $k$ be the 8-bit key. Since encryption is simply done by XORin the plaintext with the key, i.e. $e_k(m) = k \oplus m$, we have:
$$e_k(m_1) \oplus e_k(m_2) = (k \oplus m_1) \oplus (k \oplus m_2).$$
Now, the thing you need to know about XOR is that it obeys the same commutative, and associative laws as normal addition, that is:
$$a \oplus b = b \oplus a$$
$$a \oplus (b \oplus c) = (a \oplus b) \oplus c$$
and it also obeys the cancellative laws:
$$a \oplus a = \bar 0$$
$$a \oplus \bar 0 = a$$
where $\bar 0$ denotes the string of all zero bits.
Using these algebraic laws, you should be able to show that the two $k$'s on the right hand side of the equation above will cancel out, leaving only $m_1 \oplus m_2$.
Similarly, you should be able to show that, if $k \oplus m = m$, then $k = \bar 0$. The easiest way is probably to observe that, if $a = b$, than $a \oplus c = b \oplus c$. Thus, you can XOR both sides of the equation $k \oplus m = m$ with $m$, and then apply the cancellative law.