\( \begin{array}{ll}1.3 & -x^{2}+16>0 \\ 1.4 & \text { Solve for } x \text { and } y \text { simultaneously. } \\ & x-3 y=1 \text { and }(2 x+y-1)(x-y+1)=0 \\ 1.5 & \text { Prove that } \\ & \sqrt{\sqrt[3]{a b}} \cdot \sqrt[b]{b^{3} \sqrt{a}}=a^{\frac{2+b}{6 b}} b^{\frac{b+6}{6 b}}\end{array} \)

To tackle the first inequality, \(-x^{2}+16>0\), let’s rewrite it as \(x^2 < 16\). This means \(-4 < x < 4\). So, \(x\) lies between -4 and 4! It’s like finding hidden treasure on a number line; you just need to stay within those bounds! For the simultaneous equations \(x-3y=1\) and \((2x+y-1)(x-y+1)=0\), start solving the first one for \(y\): \(y=\frac{x-1}{3}\). Now plug this into either factor from the second equation, starting with \(2x+y-1=0\) to create a new equation. This leads you to coordinate pairs that represent the intersection points. Remember, it’s like putting puzzle pieces together; make sure everything fits before you declare victory!