Answer
$\qquad f(x)=2e^{\sin(x/2)}$
has absolute maxima of $2e\approx 5.43656$ at $x=3\pi+4\pi, \quad k\in \mathbb{Z}$, and
absolute minima of $2e^{-1}\approx\approx 0.73576$ at $x=\pi+4\pi, \quad k\in \mathbb{Z}$
Work Step by Step
$f_{1}(u)=e^{u}$ is an increasing function.
$(f_{1}'(u)=e^{u}\gt 0$, for all u )
Its lowest value is reached when $u $ has the lowest value,
and the highest value is reached when $u$ has the highest value.
$\left[\begin{array}{lll}
Lowest & & Highest \\
when & & when\\
& & \\
u=-1 & ...... & u=1\\
& & \\
\sin\frac{x}{2}=-1 & & \sin\frac{x}{2}=1\\
x/2=\frac{3\pi}{2}+2k\pi & & x/2=\frac{\pi}{2}+2k\pi\\
x=3\pi+4\pi & & x=\pi+4\pi
\end{array}\right]$
The function $2\cdot f_{1}(u)$ has its extrema at the same values of x.
So, $\qquad f(x)=2e^{\sin(x/2)}$
has absolute maxima of $2e\approx 5.43656$ at $x=3\pi+4\pi, \quad k\in \mathbb{Z}$, and
absolute minima of $2e^{-1}\approx\approx 0.73576$ at $x=\pi+4\pi, \quad k\in \mathbb{Z}$
