For part b, just substitute , to obtain the function value at the given point.

This is also the y-coordinate of P. So the point P is . This takes care of (i).

The gradient of the curve at P is just the function value of the derivative at P. So . I'll let you evaluate that to finish part (ii).

For part (c) use the formula of a line:

Part (b) gave you all of the information you need to fill in this formula. Does this make sense?