(a)The equation of the given curve is \\\\(y = x^2 - 2x +7\\\\).On differentiating with respect to \\\\(x\\\\), we get: \\\\(\\cfrac{dy}{dx}=2x-2\\\\)The equation of the line is \\\\(2x -y + 9 = 0.\\Rightarrow y = 2x + 9 \\\\)This is of the form \\\\(y = mx + c. \\\\)Slope of the line \\\\(= 2 \\\\)If a tangent is parallel to the line \\\\(2x- y + 9 = 0\\\\), then the slope of the tangent is equal to the slope of the line. Therefore, we have: \\\\(2=2x-2\\\\)\\\\(\\Rightarrow 2x=4\\Rightarrow x=2\\\\)Now, at \\\\(x = 2 \\\\)\\\\(\\Rightarrow y = 2^2-2\ imes 2 + 7 = 7 \\\\)Thus, the equation of the tangent passing through \\\\((2, 7)\\\\) is given by, \\\\(y-7=2(x-2)\\\\)\\\\(\\Rightarrow y -2x-3=0 \\\\)Hence, the equation of the tangent line to the given curve (which is parallel to line \\\\((2x - y + 9 = 0)\\\\) is \\\\(y-2x-3=0.\\\\)(b)The equation of the line is \\\\(5y -15x = 13. \\\\)Slope of the line \\\\(= 3 \\\\)If a tangent is perpendicular to the line \\\\(5y- 15x = 13\\\\), then the slope of the tangent is \\\\(\\dfrac{-1}{\ ext{Slope of the line}}=\\dfrac{-1}{3}\\\\).\\\\(\\Rightarrow \\dfrac{dy}{dx}= 2x-2=\\dfrac{-1}{3}\\\\)\\\\(\\Rightarrow 2x =\\dfrac{-1}{3}+2\\\\)\\\\(\\Rightarrow 2x=\\dfrac{5}{3}\\\\)\\\\(\\Rightarrow x=\\dfrac{5}{6}\\\\)Now, at \\\\(x=\\dfrac{5}{6}\\\\)\\\\(\\Rightarrow y=\\dfrac{25}{36}-\\dfrac{10}{6}+7=\\dfrac{25-60+252}{36}=\\dfrac{217}{36}\\\\)Thus, the equation of the tangent passing through \\\\(\\left ( \\dfrac{5}{6}, \\dfrac{217}{36} \\right )\\\\) is given by, \\\\(\\left(y-\\dfrac{217}{36} \\right )=-\\dfrac{1}{3}\\left ( x-\\dfrac{5}{6} \\right )\\\\)\\\\(\\Rightarrow =\\dfrac{36y-217}{36}-\\dfrac{1}{18}(6x-5)\\\\) \\\\(\\Rightarrow 36y-217=-2(6x-5)\\\\)\\\\(\\Rightarrow 36y-217=-12x+10\\\\)\\\\(\\Rightarrow 36y+12x-227=0\\\\) Hence, the equation of the tangent line to the given curve (which is perpendicular to line \\\\(5y-15x = 13)\\\\) is \\\\(36y+12x-227=0\\\\)