= Solution
Put $h=x-\pi/2$. The <sine> addition identity gives $\sin x=\cos h=1-h^2/2+h^4/24+O(h^6)$. Applying the <exponential series> gives
$$
\begin{aligned}
e^{\sin x}&=e\exp\left(-\frac{h^2}{2}+\frac{h^4}{24}+O(h^6)\right)\\
&=e\left[1-\frac{h^2}{2}+\left(\frac1{24}+\frac18\right)h^4+O(h^6)\right].
\end{aligned}
$$
Thus the first three nonzero terms in the <Taylor series> about $\pi/2$ are
$$
\boxed{e^{\sin x}=e-\frac e2\left(x-\frac\pi2\right)^2+\frac e6\left(x-\frac\pi2\right)^4+O\left(\left(x-\frac\pi2\right)^6\right).}
$$
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