The integral is
$$\int _{|z-\pi/4|=\pi/8} \frac{e^{\sin z}}{z^2(z-\pi/4)}dz$$
I assume the pole is located at $(\pi/4,0)$ with a radius of $\pi/8$.
My attempt:
I drew a circle centered at $(0,0)$ with a radius with of $\pi/8$ and vice versa with $(\pi/4,0)$ and pinched it?
I then, splitted it up.
$$=\int _{c_1} \frac{\frac{e^{\sin z}}{z-\pi/4}}{z^2}dz + \int _{c_1} \frac{\frac{e^{\sin z}}{z^2}}{z-\pi/4}dz$$
and I end it up getting an expression of,
$\frac{e^{1/\sqrt{2}}}{32\sqrt{2}\pi^2}(-8\pi^2\sqrt2-\pi-4\sqrt2)$
and the answer is $\frac{32e^{1/\sqrt{2}}i}{\pi}$
Considering that this is about Cauchy's integral formula, you should simply notice that your integral is equal to
$$\int_C \frac{e^{\sin z} / z^2}{z - \pi/4} \, dz = 2 \pi i \frac{e^{\sin z}}{z^2} \big|_{z = \pi/4}$$
since $e^{\sin z}/{z^2}$ is holomorphic away from $0$. You can easily simplify this to get
$$\frac{32i}{\pi} e^{\sqrt{2}/2}$$
which is the desired answer.