For the first one, set $\sqrt[6]{x} = y$. This means $\sqrt{x} = y^3$ and $\sqrt[3]{x} = y^2$. We then get that $\lim_{x \to 1} \dfrac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{y \to 1} \dfrac{y^3-1}{y^2-1} = \lim_{y \to 1} \dfrac{(y-1)(y^2+y+1)}{(y-1)(y+1)} = \lim_{y \to 1} \dfrac{(y^2+y+1)}{(y+1)} = \dfrac32$ For the second one, as the hint suggests, let $y = \pi - x$. We then get that \begin{align} \lim_{x \to \pi} \dfrac{\sin(5x)}{\sin(3x)} & = \lim_{y \to 0} \dfrac{\sin(5(\pi-y))}{\sin(3(\pi-y))} = \underbrace{\lim_{y \to 0} \dfrac{\sin(5 \pi -5y)}{\sin(3\pi-3y)} = \lim_{y \to 0} \dfrac{\sin(5y)}{\sin(3y)}}_{(\star)}\\ & = \lim_{y \to 0} \dfrac{5 \times \dfrac{\sin(5y)}{5y}}{3 \times \dfrac{\sin(3y)}{3y}} = \dfrac{5 \times \lim_{y \to 0} \dfrac{\sin(5y)}{5y}}{3 \times \lim_{y \to 0} \dfrac{\sin(3y)}{3y}} = \dfrac53\\ \end{align} where $(\star)$ follows from the fact that $\sin((2n+1)\pi-y) = \sin(y)$ and the last equality comes from the fact that $\displaystyle \lim_{\theta \to 0} \dfrac{\sin(\theta)}{\theta} = 1$.
EDIT
$\sin((2n+1)\pi-y) = \overbrace{\sin(2n \pi + \pi -y) = \sin(\pi -y)}^{\text{Because $\sin$ is $2 \pi$ periodic.}}$ Now $\sin(\pi-y) = \sin(\pi) \cos(y) - \cos(\pi) \sin(y) = 0 - (-1) \sin(y) = \sin(y)$