Let $P\left(x_{1}, y_{1}\right)$ be the required point. The given curve is
$$
\begin{aligned}
y &=6 x-x^{2} \\
\Rightarrow \quad \frac{d i j}{d x} &=6-2 x \\
\Rightarrow \quad\left(\frac{d y}{d x}\right)_{\left(x_{1}, \mathfrak{h}\right)} &=6-2 x_{1}
\end{aligned}
$$
Since, the tangent at $\left(x_{1}, y_{1}\right)$ is parallel to the line $4 x-2 y-1=0$.
$\therefore$ Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the line $4 x-2 y-1=0$
$\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(x, b)}=2$
$\Rightarrow \quad 6-2 x_{1}=2 \Rightarrow x_{1}=2$
$\therefore \quad y_{1}=6 x_{1}-x_{1}^{2}=6 \times 2-2^{2}=8$
Thus, required point is $(2,8)$.