Yes, you are over complicating the question. You don't need to pick out 'point R' when you can get the information through much simpler working out (which is shown in the solutions from Insight).
Picking out Point R is kind of a good idea, I guess, but you've ignored some key information regarding the motion, hence why your answer is incorrect. It so happens that we aren't given enough details to figure out this key information, which means that we can't use this 'point R' method.
For the second picture, I figured that at point R, the vertical velocity component would be the same as the inital vertical velocity at the top, right corner of the tower.
Not necessarily. You're ignoring the fact that a vertical acceleration will act upon the motorbike for a period of time - a period of time. This will mean the the vertical velocity at point R will be different. You could think of it this way too: the velocity throughout the motion will decrease, the horizontal velocity remains constants, therefore the vertical velocity must be the value that decreases.
The other thing:
Then I let 
,
,
,
Why'd you let vertical velocity at this point be 0? We know that the bike is still in motion for a bit yet, so if vertical velocity is 0, (and hence gravitation force being ignored too) then it'd only have a velocity horizontally. This would mean that the bike would travel horizontally along the air, it wouldn't drop. That clearly doesn't happen.
Another thing is that you made a calculation mistake here:
Which would be )
x 
=
.
 = \frac{\sqrt{3}}{2})
, not 1/2
(which equals whatever when you round it off to whatever decimal places)
Also I will point out is that if anything, point R would be at the end of travelling distance 'R' - i.e. the end of the motion. (irrelevant though, since you could have used any other letter).
So let's try your point R method, taking into account all information we have.
Looking at vertical information only, so height is given by:
We need to figure out how long it takes to reach point R, knowing that
and that
We can't use the vertical values, because the vertical value of 'x' at point R is what we're trying to figure out in the first place.
We can figure the time from the horizontal values:
Oh no, we can't do this, we don't know quantitatively what the horizontal displacement is. Nor are we given enough information about it - we're only told that the bike travels a distance 'R'. So since time and displacement at point R are unknown, we don't have enough information to figure out more details than what we're given using any of the equations of motion.
So another method is needed, let's try what you tried to do:
We have u, we have a, we're trying to figure out x... but what is v? We don't have enough information.
But what if we considered the horizontal values, we know v will remain constant horizontally (as there is no air resistance etc.) and perhaps we can figure out what 'x' is and then figure out the time and other information we need.
 = \frac{\sqrt{3}}{2})
^2 = (\frac{\sqrt{3}}{2})^2 + 2ax)
So what's the horizontal acceleration? It's 0, isn't it - since there's no forces acting in this direction and that the velocity remains constant (newton's first law). This reason is also why we can pretty much just use this equation for horizontal components:
from which we can also quickly see that we have two unknowns.
So I hope I was able to point out what you did wrong, I'm sure you'll be able to deduce the correct method of working out for this question yourself.
Edit: fixed a few mistakes, misinterpreted something, doesn't affect main point, but see my next post for clarification
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