Showing a function can be continuously extended to the unit circle Fix $0 < a < \infty.$ Define for $|z| < 1$ the function $$ f(z) = \sum_{n=0}^\infty 2^{-na} z^{2^n}$$
and show that $f$ extends continuously to the unit circle but can not be analytically continued past the unit circle. 
The second part of the problem is a whole new can of worms but I'm even having difficulty with the first part: it's easy to show that $|f|$ is bounded if $z = e^{i\theta}$ but I can't quite figure out the continuity piece. 
Let $z_0 = re^{i\phi}$ be a point in the unit disk. Then 
$$f(z) - f(z_0) = \sum_{n=0}^{\infty} 2^{-na}(e^{i\theta 2^n} - r^{2^n} e^{i\phi 2^n}) 
\\= \sum_{n=0}^\infty 2^{-na}[(1+r^{2^n})(e^{i\theta 2^n} - e^{i\phi2^n}) + 
(e^{i\phi 2^n} - r^{2^n} e^{i\theta 2^n})]$$
and I tried finding some bounds for $|f(z) - f(z_0)|$ given bounds for $|z-z_0|$ in this manner (and some other ways) but I wasn't able to find any results. 

if |z|$\leqslant$1 then the sum of the magnitudes of your series is dominated by $\sum_{n=0}^\infty$2^-na $\leqslant$ $\frac1{1-2^ --a}$ 
This shows by the Weierstrass Majorant test that your series converges uniformly to a continuous function for |z|$\leqslant$1  .This settles the first part of your problem .
