reply to the first qn .

Take y to be constant :

differentiating both sides wrt x , we get f'( x + y^n ) = f'(x ) for all y

which means f'( x ) = constant = c ( let )

so f( x ) = cx + d  ( d = constant )

Now putting x = y = 0 in the given reln we get that f(0) = 0

giving , d= 0

so f( x ) = cx

Now put x = 0  in the given relation , so we get

f( y^n ) = [f(y )]^n

so cy^n = c^n y^n for all y

giving , c= 1

so f( x ) = x  ; thus f(5 ) + f'(5 ) = 5 + 1 = 6