I'm taking "part 3" to mean the 3rd of the posted questions.
If [tex]f(x)=(x^3+2x)\ln x[/tex], then
[tex]f'(x)=(x^3+2x)'\ln x+(x^3+2x)(\ln x)'=(3x^2+2)\ln x+\dfrac{x^3+2x}x[/tex]
[tex]\implies f'(x)=(3x^2+2)\ln x+x^2+2[/tex]
Then when [tex]x=1[/tex], we get [tex]f'(1)=3[/tex].
