Cambridge's List of Formulas gives you the curved surface area of a cylinder and cone, the surface area of a sphere, and the volume of a prism, pyramid, cylinder, cone and sphere:

You are not given the formula for total surface area of a cuboid or the flat circular ends of a cylinder/cone — you build these from area facts you already know (e.g. total surface area of a cylinder = curved surface area + area of the two circular ends).

The syllabus defines a prism as any solid with a uniform (identical, unchanging) cross-section running through its length — this includes shapes like a triangular prism, but also less obvious cases such as a cylindrical sector (a "wedge" cut from a cylinder). Recognising the constant cross-section is what tells you to use $V = \text{area} \times \text{length}$ rather than a cone/pyramid-style formula.

The slant height, the radius, and the vertical height of a cone form a right-angled triangle, so:

This is needed constantly for curved-surface-area questions, since the curved surface area formula uses $l$, not the vertical height $h$.

Similar reasoning applies to pyramids: the vertical height, the distance from the centre of the base to the midpoint of a side (for the slant height of a triangular face), or the distance to a base vertex (for a slant edge), together form right-angled triangles you can solve with Pythagoras. Sketching the specific right-angled triangle hidden inside the 3D solid — rather than trying to picture the whole pyramid at once — makes it far easier to see which two lengths you already know and which one you're solving for.

The $\dfrac13$ factor appears only for pyramids and cones — solids that taper to a single point (an apex). A prism or cylinder has the same cross-sectional area all the way along its length, so no such factor is needed; a pyramid or cone's cross-section shrinks steadily to zero at the apex, and it's a geometric fact (provable using calculus, though you don't need to prove it here) that this tapering means the volume is exactly one third of the equivalent prism/cylinder with the same base and height.

A cone has base radius $5\text{ cm}$ and vertical height $12\text{ cm}$. Find its total surface area in terms of $\pi$.

1. Find the slant height: $l = \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}$.
2. Curved surface area: $\pi r l = \pi(5)(13) = 65\pi\text{ cm}^2$.
3. Area of the circular base: $\pi r^2 = \pi(5)^2 = 25\pi \text{ cm}^2$.
4. Total surface area $= 65\pi + 25\pi = 90\pi \text{ cm}^2$.

A pyramid has a square base of side $10\text{ cm}$ and a slant edge (apex to a base corner) of $13\text{ cm}$. Find the volume of the pyramid.

1. The base diagonal is $\sqrt{10^2+10^2} = \sqrt{200}\text{ cm}$, so half the diagonal (base centre to a corner) is $\dfrac{\sqrt{200}}{2}\text{ cm}$.
2. The vertical height, half-diagonal, and slant edge form a right-angled triangle: $h = \sqrt{13^2 - \left(\tfrac{\sqrt{200}}{2}\right)^2} = \sqrt{169 - 50} = \sqrt{119} \approx 10.9\text{ cm}$ (3 s.f.).
3. Volume $= \tfrac13 \times \text{base area} \times h = \tfrac13 \times 100 \times 10.9 \approx 364\text{ cm}^3$ (3 s.f.).