#### Answer

$$1.14 \times 10^{-5}$$

#### Work Step by Step

Since
\begin{array}{ll}
{f(x)=e^{\sin x},} & {f\left(\frac{\pi}{2}\right)=e} \\
{f^{\prime}(x)=e^{\sin x} \cos x,} & {f^{\prime}\left(\frac{\pi}{2}\right)=0} \\
{f^{\prime \prime}(x)=e^{\sin x} \cos ^{2} x-e^{\sin x} \sin x,} & {f^{\prime \prime}\left(\frac{\pi}{2}\right)=-e}
\end{array}
Then
\begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=e+(0)\left(x-\frac{\pi}{2}\right)-\frac{e}{2}\left(x-\frac{\pi}{2}\right)^{2} \\
&=e-\frac{e}{2}\left(x-\frac{\pi}{2}\right)^{2}
\end{aligned}
and
$$\left|f(1.5)-T_{2}(1.5)\right|=|2.711481018-2.711469651|=1.14 \times 10^{-5}$$