Second one :D
(2^x)(e^{\frac{1}{\sin x}})
\(f(x)=(2^x)(e^{\frac{1}{sin x}})\\ f(x)=(2^x)e^{(sinx)^{-1}}\\ f(x)=uv\;\;\;where\;\;\;u=2^x\;\;\;and\;\;\;v=e^{(sinx)^{-1}}\\~\\ v=e^{(sinx)^{-1}}\\ \frac{dv}{dx}=-(sinx)^{-2}(cosx)e^{(sinx)^{-1}}\\ \frac{dv}{dx}=\frac{-(cosx)e^{(sinx)^{-1}}}{(sinx)^{2}}\\~\\ u=2^x\\ ln(u)=ln(2^x)\\ ln(u)=xln(2)\\ x=\frac{ln(u)}{ln2}\\ \frac{dx}{du}=\frac{1}{u*ln2}\\ \frac{du}{dx}=u*ln2\\ \frac{du}{dx}=2^x*ln2\\ \)
\(f(x)=(2^x)(e^{\frac{1}{sin x}})\\ f(x)=(2^x)e^{(sinx)^{-1}}\\ f(x)=uv\;\;\;where\;\;\;u=2^x\;\;\;and\;\;\;v=e^{(sinx)^{-1}}\\~\\ v=e^{(sinx)^{-1}}\\ \frac{dv}{dx}=\frac{-(cosx)e^{(sinx)^{-1}}}{(sinx)^{2}}\\~\\ u=2^x\\ \frac{du}{dx}=2^x*ln2\\ \mbox{Using product rule}\\ f'(x)=2^x*\;\frac{-(cosx)e^{(sinx)^{-1}}}{(sinx)^{2}}\;\;+\;\;e^{(sinx)^{-1}}\;*\;2^xln2\\ f'(x)=2^x*\;\frac{-(cosx)e^{cosec(x)}}{(sinx)^{2}}\;\;+\;\;e^{cosec(x)}\;*\;2^xln2\\ f'(x)=2^xe^{cosec(x)}\left(\;\frac{-(cosx)}{(sinx)^{2}}\;\;+\;\;ln2\right)\\ f'(x)=\frac{2^xe^{cosec(x)}}{sin^2x}\;(ln2\;sin^2(x)\;-\;cosx)\\ \)
My answer is the same as Wolfram Alpha although it is arranced a little differently
http://www.wolframalpha.com/input/?i=differentiate++2%5Ex*e%5E(1%2Fsinx)