Question 1118084
Since there is only one blue one:
P(1 blue and 1 red) = 1 - P(both red) = {{{ 1 - (n/(n+1))*((n-1)/(n)) }}} 

= {{{ 1 - ((n-1)/(n+1)) }}}
= {{{  (n+1)/(n+1) - (n-1)/(n+1) }}} 
=  {{{ highlight( 2/(n+1) ) }}}

So,  to sanity check  this:
Say n=1:  P(1 blue and 1 red) = 2/(1+1) = 1  (ok, we must get one of each)
If n=2:  P(1 blue and 1 red) = 2/(2+1) = 2/3   (ok, b/c 1/3 of the time you'd expect two reds) 
If n=100:  P(1 blue and 1 red) = 2/(100+1) = 2/101  (seems ok, very unlikely to happen given all those reds, much more likely to get two reds)<br>

To do part(b), set the equation above equal to 0.125 (= 1/8)  then solve that equation for n.