Why is this number so close to $1$? The only positive solution of the equation $\sin (\tan x) = x$ is at a number $a = 0.999906...$. Is it a coincidence that the number $a$ is so close to $1$, or is there a conceptual explanation?
It was obvious $a$ was going to have to be less than $1$, but not this close. The answer seems to be related to $\tan 1 \approx \pi/2$, but that merely shifts the problem to explaining why that is the case. 
(I became interested in this question after reading this other question: Prove: $\sin (\tan x) \geq {x}$ )
Edit: I have thought about the question some more, and added some ideas in the form of an answer partially addressing why we have $\tan 1 \approx \pi/2$.
 A: J Swanson and Christian Blatter's answers address the question of why, given that $\tan 1 \approx \pi/2$ (within $1 \%$), we have $\sin \tan 1 \approx 1$ within $0.01\%$. Taylor's formula for $\sin x$ about $x = \pi/2$ shows why the accuracy would be approximately squared.
That leaves open the question of why $\tan 1 \approx \pi/2$. 
Gauss's continued fraction for $\tan z$ (see http://en.wikipedia.org/wiki/Gauss_continued_fraction ) is:
$$\tan z = \cfrac{z}{1 - \cfrac{z^2}{3 - \cfrac{z^2}{5 - \cfrac{z^2}{7 - {}\ddots}}}}.$$
In particular, for $z = 1$, this yields
$$ \tan 1 \approx  \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5}}} = \frac{14}{9}.$$
This is an excellent approximation, valid to within just over $0.1\%$. So we need only consider the question of why $\pi \approx 28/9$. 
I don't have a very good answer for this part, as I haven't been able to locate any standard approximating sequences of $\pi$ one of whose first terms is $28/9$. However, you can obtain $28/9$ as an approximation of $\pi$ in the following way, as explained in A History of Pi by Beckmann. (If the circle has area $\pi$, then the octagon has area $28/9$.)

A: It happens that $\pi/2 - \tan 1 \approx 0.0133886$. This may just be a numerical coincidence or there may be a good reason. But in any case, $\sin$ is very flat close to $\pi/2$, so even values that are just pretty close to $\pi/2$ get translated to values very close to $1$, hence $\sin \tan 1 \approx 0.9999103$. By continuity/smoothness the fixed point must be very close to $1$, as you say.
I'll make this more quantitative in an effort to make it more "explanatory". Suppose $f(x)$ is smooth near $x=1$ and $f(1) - \pi/2 = \delta$. Then $\sin(f(1)) = \cos(\delta) = 1 - \delta^2/2 + \cdots$. Hence $\sin f(x) - x$ at $x=1$ is $-\delta^2/2 + \cdots$. If $|\delta| \approx 0$, the difference is approximately $0$, so if a fixed point exists, it should be near here, which can be made rigorous by comparing the derivative of $\sin f(x) - x$ to $-\delta^2/2 + \cdots$ at $x=1$.
In the present case with $f = \tan$, we find $\delta \approx -0.0133886$, we have $-\delta^2/2 \approx -0.0000896273$, and this is tiny compared to $(\sin \tan x - x)'(1) \approx -0.954138$, so we'll definitely get a fixed point very close to here. Indeed, the actual fixed point differs from $1$ by $0.0000939875$, which is very close to my estimate, and moreover $0.0000896273/0.954138 \approx 0.0000939354$. That is, the closeness of the fixed point to $1$ is just an amplified version of the closeness of $\tan 1$ to $\pi/2$ thanks to the sine function.
Some thoughts on $\pi/2 - \tan 1 \approx 0.0133886$: we can change the problem in at least two ways--replace $\sin$, $\pi/2$, $1$; replace $\tan$--and sometimes the above reasoning will still go through. There are a lot of potential tweaks, so I'd be satisfied by the coincidence explanation. On the other hand, it would be more interesting if was something really was going on with $\tan$ and $\pi/2$.
A: Consider that you try to solve your equation using Newton method starting at $x=1$. The first iterate is given by 
$$x = 1 + \frac {-1 + \sin(\tan(1))} {1 - \cos(\tan(1)) \sec(1)^2};$$
the change is $-0.000093934$.   
If you use Halley method, the formula becomes much more complex and the change is $-0.0000939875$.
A: From
$${4\over3}=1+{(1)^3\over3}<\tan 1<\tan{\pi\over3}=\sqrt{3}$$
it follows that
$$\tan 1={\pi\over2}+\tau$$ for a $\tau$ of small absolute value. That in fact $\tau\doteq-0.0134$ and therefore $|\tau|\ll1$ is pure luck.
For a $\delta$ with $0<\delta\ll 1$ we therefore have
$$\tan(1-\delta)\doteq{\pi\over 2}+\tau-\delta\left(1+{\pi^2\over4}\right)\doteq {\pi\over2}+\tau-{7\over2}\delta\ ,$$
where we have made use of $\pi^2\doteq10$. It follows that
$$\sin\bigl(\tan(1-\delta)\bigr)\doteq 1-{1\over2}\left(\tau-{7\over2}\delta\right)^2\ ,$$
and this should be equal to $1-\delta$ for some positive $\delta\ll1$. The equation
$$\left(\tau-{7\over2}\delta\right)^2=2\delta$$
has two solutions $\delta_1\doteq{\tau^2\over2}$ and $\delta_2\doteq{8\over49}$, the first of which is the one we are interested in. Given the value of $\tau$ from above we obtain $\delta_1\doteq0.0000896$, which leads to $x=0.9999103$.
