Solve the following inequalities:

$$\begin{gathered} x^2-12x+27<0 \\ \Rightarrow x^2-9x-3x+27<0 \\ \Rightarrow x\left(x-9\right)-3\left(x-3\right)<0 \\ Here, if we take x<3 \\ Then, \left(x^2-12x+27\right)<0 will not be true and if we take x<9. \\ So, x must lie between 3<x<9, \\ Because for each value between 3<x<9, (x^2-12x+27)<0 will be true. \\ \mathit{a}\mathit{n}\mathit{s}. 3 <x<9. \end{gathered}$$

$$\begin{gathered} x^2-13x+40\ge 0 \\ \Rightarrow x^2-8x-5x+40\ge 0 \\ \Rightarrow x(x-8)-5(x-8)\ge 0 \\ \Rightarrow (x-8) (x-5)\ge 0 \\ Here, the product of (x-8) and (x-5) is grater than zero or equal to zero. \\ So the value of these must be both positive or negative. \\ When both positive, (x-8)\ge 0 \\ \Rightarrow x>8; (x-5)\ge 0 \\ \Rightarrow x\ge 5 [It is not true] \\ When both negative, -(x-8)\ge 0 \\ \Rightarrow x\le 8; [It is not true] \\ -(x-5)\ge 0 \\ \Rightarrow x\le 5; \\ So, the range x\ge 8 Or, x\le 5. \end{gathered}$$