Probability that second ball is magenta 
attempt
Notice that in any urn, we have $r-1+n-r = n-1$ balls. We have to pick 2 balls, so the sample space size is ${n-1 \choose 2 }$. Now if we want the second ball to be mangenta, then we must have either $MM$ or $RM$. So the probability of this is
$$ P = \frac{ MM + RM }{{n-1 \choose 2} } = \frac{ \frac{(n-r)^2}{(n-1)^2} + \frac{r-1}{n-1} \cdot \frac{n-r}{n-1} }{{n-1 \choose 2}} $$
but the answer the lecturer gives is $1/2$, does the above answer seems reasonable?
Now, for second part, we want $P(MM)$ which is just
$$ \frac{ \frac{(n-r)^2}{(n-1)^2}  }{{n-1 \choose 2}} $$
is this correct?

In the first urn there are $(r - 1)$ red balls i.e. $0$ red balls and $(n-r)$ i.e. $(n-1)$ magenta balls. There are two ways the second ball drawn will be magenta $RM$ and $MM$. So, the number of ways to select that (the first term in the square brackets shows for the number of favorable ways of $RM$ and the second one reflects for $MM$) is $[0 + \binom{n-1}{1}* \binom{n-2}{1}]$. 
Similarly for the second urn $[\binom{1}{1}*\binom{n-2}{1} + \binom{n-2}{1}*\binom{n-3}{1}]$ ways. So, we end up with this sequence:
$[0 + \binom{n-1}{1}* \binom{n-2}{1}]$ - first urn
$+$ $[\binom{1}{1}*\binom{n-2}{1} + \binom{n-2}{1}*\binom{n-3}{1}]$  - second urn
$+$ $[\binom{2}{1}*\binom{n-3}{1} + \binom{n-3}{1}*\binom{n-4}{1}]$ - third urn
$+...+[\binom{n-3}{1}*\binom{2}{1} +  \binom{2}{1}*\binom{1}{1}]$ - third from last urn
$+ [\binom{n-2}{1}*\binom{1}{1} + 0 ] $ - second from last
$+ 0$. - last urn
Note that $0$ is for the last urn which does not have any magenta balls.
Adding all of the above terms will give:
$[(n-2)*\frac{n*(n-1)}{2}]$. And the total number of ways the one urn can be selected is $n$. And selecting two balls without replacement will be $(n-1)*(n-2)$ ways. So, the probability will be:
$[(n-2)*\frac{n*(n-1)}{2}]$/$n*(n-1)*(n-2)$ = $0.5$
