A curve has equation y=ax^3+bx^2+cx+d

where a,b,c and d are constants. The curve passes through the points (0,-6) and (1,-8). At these two points the curve has gradients -5 and 2, respectively.
a) Find the values of a,b,c and d.
b)Show that the curve crosses the x axis at point (2,0)
c) Find the coordinates of the other two points where the curve crosses the x-axis

The two gradients has mixed me up and i am not sure how to approach it.

By gradient do you means slope? I'll assume you do.

d/dx ax^3+bx^2+cx+d = 3ax^2+2bx^+c = y'(x)

y'(0) = 3a0^2+2b0^+c = -5

c = -5

y'(1) = 3a1^2+2b1^+ c = 2

y(-5) = y=a-5^3+b-5^2+-c-5+d = 0

y(2) = y=a2^3+b2^2+c2+d = 0

You now have four equations and four unknowns so you can solve for the four unknowns.

a=1
b=2
c=-5
d=-6

b should be easy as you just need to show the given the input of 2 you get 0 as the output as you expect.
y(2) = 1*2^3+2*2^2+-5*2 - 6 = 0

To do c you just set the equation to 0 and solve. You should get three solutions.

x^3+2x^2+-5x+-6 = 0

factoring you get
x^3+2x^2+-5x+-6 = (x - 2)(x + 1)(x + 3)
therefore
y(x) = 0 when x = 2, -1, -3

y=ax^3+bx^2+cx+d

(0,-6) -5
(1,-8) 2

y'=3ax^2+2bx+c

-5=3a(0)^2+2b(0)+c
2=3a(1)^2+2b(1)+c
-6=a(0)^3+b(0)^2+c(0)+d
-8=a(1)^3+b(1)^2+c(1)+d

c=-5
3a+2b+c=2
d=-6
a+b+c+d=-8

put eq1 and eq3 into equ2 and equ3

3a+2b-5=2
a+b-5-6=-8

3a+2b=7
a+b=3

a=1
b=2

so

y=x^3+2x^2-5x-6

now divde (x^3+2x^2-5x-6)/(x-2)
x^2+4x+3
(x+3)(x+1) so other roots are x=-1 and x=-3

a)

y' = 3ax^2 + 2bx + c

At x=0:

y(0)=-6 ==> -6 = d
y'(0)=-5 ==> -5 = c

At x=1:

y(1)=-8 ==> -8 = a + b + c + d = a + b -11 ==>a+b = 3

y'(1)=2 ==> 2 = 3a + 2b + c = 3a + 2b - 5 ==> 3a+2b = 7

a = 1, b = 2

y = x^3 + 2x^2 - 5x - 6

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b)

y(2) = 8 + 8 - 10 - 6 = 0

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c) x^3 + 2x^2 - 5x - 6 = x^3 - 2x^2 + 4x^2 - 8x + 3x - 6 = (x-2)(x^2+4x+3) = 0

x = -1, x = -3

first dy/dx= 3ax² +bx + c

so 3a(-2)² +b(-2) + c = 0
and 3a(2)² +b(2) + c = 0