Your working for c) i) and ii) is correct, however, when finding values of 4θ, it is a better practice to write the solutions as 4θ = ±π/2, ±3π/2, ... (using i) as an example) as the solutions will occur in more quadrants than one. Once plugging in values for θ, you should also recognise that the expressions from c) i), ii) and iii) are also even functions, meaning that the ± sign should appear in front of the cos. Additionally, since the expressions are of degree 4, there can only be a maximum of 4 roots.
For c) iii), use the same method you used for ii) to obtain cos4θ = -1/2 and θ = 2π/3, 4π/3 then once again solving in each quadrant for θ and plugging back into a value for x, which will have exact values you can evaluate this time.
For c) iv), assuming you have already derived the expression for tan4θ in terms of tanθ, moving the 4x^3 - 4x term to the other side will yield 1-6x^2+x^4 = 4x-4x^3. Dividing both sides by 1-6x^2+x^4 to get 1 = (4x-4x^3)/(1-6x^2+x^4) and noticing the similarity between this expression and the expression for tan4θ, you can apply the substitution x = tanθ. You should end up with the result 1 = tan4θ.
Solving for θ; 4θ = π/4, 5π/4, -3π/4, -7π/4 ⇒ θ = π/16, 5π/16, -3π/16, -7π/16, and finally using the identity tan(-x) = -tanx, allows for tan(-3π/16) and tan(-7π/16) to be written as -tan(3π/16) and -tan(7π/16), respectively.
