Use the Contraction Mapping Principle to show that $x=\frac19\sin\left(3x\right) + \sqrt{x}$ has exactly one solution $x\geqslant\frac{8}{9}$ 
Use the Contraction Mapping Principle to show that $x=\frac19\sin\left(3x\right) + \sqrt{x}$ has exactly one solution $x\geqslant\frac{8}{9}$.

I have literally no idea if this is right, please could someone check my answer? 
Let $T$ be defined by $Tx=\frac19\sin\left(3x\right) + \sqrt{x}$. To prove that $T$ is a contraction:
$|Tx-Ty| = \left|\frac{\sin(3x)}{9} + \sqrt{x} - \frac{\sin(3y)}{9} - \sqrt{y}\right| \le \max\left|\frac {\ 3cos(3c)}9 + \frac 1{2\sqrt{c}}\right| \times |x-y|$
by the Mean value theorem
$= \left(\frac{1}{2}+\frac{1}{3}\right) |x-y| = \frac{5}{6} |x-y| $
$\lambda = 5/6 $ 
Which proves the Contraction (although I haven't shown that the metric space $X$ is complete! How would I do this?)
Using iterative methods, with $x_0 = 1$, the fixed point converges quite quickly to $1.01877$.
Thank you!

$[\frac 8 9,\infty)$ is a closed subset of a complete metric space ($\mathbb R)$,  hence it is complete.   Simple proof that closed subsets of complete spaces are complete:   The space itself is complete, so any cauchy sequence in the subspace has a limit in the space itself. But then it's a convergent sequence in the big space,  so since the subset is closed, it contains all of its limit points.
