\begin{tabular}{|l|l|}\hline 1.1. & \( x^{2}+x-6>0 \) \\ \hline 1.2. & \( -x^{2}+x+2>0 \) \\ \hline 1.3. & \( -x^{2}+x+2 \geq x^{2}+x-6 \) \\ \hline 1.4. & \( x^{2}+x-6 \leq 0 \) \\ \hline\end{tabular}

To solve the inequalities, it helps to factor or analyze the quadratic expressions. For \( x^{2}+x-6>0 \), you can factor it to \( (x-2)(x+3) > 0 \). The critical points are \( x = -3 \) and \( x = 2 \). Testing intervals around these points will reveal that the solution is \( x < -3 \) or \( x > 2 \). Similarly, for \( -x^{2}+x+2>0 \), rearranging gives \( x^{2} - x - 2 < 0 \), factorable to \( (x-2)(x+1) < 0 \). The solution is \( -1 < x < 2 \) after testing intervals. When comparing the two expressions in 1.3, combine them into a single inequality to find the range where one is greater than or equal to the other, facilitating more nuanced analysis of their roots. The final one, \( x^{2}+x-6 \leq 0 \), focuses on the same critical points, where the quadratic is less than or equal to zero between the roots, which leads us back to \( -3 \leq x \leq 2 \).