Answer
$$\eqalign{
& {\bf{a}}.\,y = 10x - 20 \cr
& {\bf{b}}.y = 12x - 33 \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}g\left( x \right) = {x^2} + f\left( x \right) \cr
& {\bf{a}} \cr
& g\left( 3 \right) = {\left( 3 \right)^2} + f\left( 3 \right) \cr
& g\left( 3 \right) = 9 + 1 \cr
& g\left( 3 \right) = 10 \cr
& {\text{Point }}\left( {3,10} \right) \cr
& \cr
& {\text{Calculate }}g'\left( 3 \right) \cr
& g'\left( x \right) = 2x + f'\left( x \right) \cr
& g'\left( 3 \right) = 2\left( 3 \right) + f'\left( 3 \right) \cr
& g'\left( 3 \right) = 6 + 4 \cr
& g'\left( 3 \right) = 10 \cr
& {\text{The equation of the tangent line is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 10 = 10\left( {x - 3} \right) \cr
& y - 10 = 10x - 30 \cr
& y = 10x - 20 \cr
& \cr
& {\bf{b}}. \cr
& {\text{Let }}h\left( x \right) = 3f\left( x \right) \cr
& h\left( 3 \right) = 3f\left( 3 \right) \cr
& h\left( 3 \right) = 3\left( 1 \right) \cr
& h\left( 3 \right) = 3 \cr
& {\text{Point }}\left( {3,3} \right) \cr
& \cr
& {\text{Calculate }}h'\left( 3 \right) \cr
& h'\left( x \right) = 3f'\left( x \right) \cr
& h'\left( 3 \right) = 3f'\left( 3 \right) \cr
& h'\left( 3 \right) = 3\left( 4 \right) \cr
& h'\left( 3 \right) = 12 \cr
& {\text{The equation of the tangent line is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 3 = 12\left( {x - 3} \right) \cr
& y - 3 = 12x - 36 \cr
& y = 12x - 33 \cr} $$