Answer
$$z=\frac{1}{2}+ \frac{\sqrt{3}}{2} (u-\pi/6)+ \frac{\sqrt{3}}{12}(w-1)$$
Work Step by Step
Given $$G(u, w)=\sin (u w), \quad\left(\frac{\pi}{6}, 1\right)$$
Since
\begin{align*}
G(u, w)&=\sin (u w)\ \ \ \ &G\left(\frac{\pi}{6}, 1\right)&=\frac{1}{2} \\
G_u(u, w)&=w\cos(uw) \ \ \ \ &G_u \left(\frac{\pi}{6}, 1\right)&= \frac{\sqrt{3}}{2}\\
G_w(u, w)&= u \cos(uw)\ \ \ \ & G_w\left(\frac{\pi}{6}, 1\right)&= \frac{\sqrt{3}\pi}{12} \\
\end{align*}
Then, the tangent plane at $(1,0.8)$ is given by
\begin{align*}
z&=G\left(\frac{\pi}{6}, 1\right)+G_u \left(\frac{\pi}{6}, 1\right)(u-\pi/6)+G_v \left(\frac{\pi}{6}, 1\right)(w-1)\\
&=\frac{1}{2}+ \frac{\sqrt{3}}{2} (u-\pi/6)+ \frac{\sqrt{3}}{12}(w-1)
\end{align*}
