\(x_{n+1} = \dfrac{(x_n)^3-1}{4},~x_1 = -1\\ x_2 = \dfrac{(x_1)^3-1}{4} = \dfrac{(-1)^3-1}{4} = -\dfrac 1 2\\ x_3 = \dfrac{(x_2)^3-1}{4} = \dfrac{\left(-\dfrac 1 2\right)^3 - 1}{4} = -\dfrac{9}{32}\)
\(\text{The solution occurs when the recursion has stabilized on a single value}\\ \text{so just run the iteration until }x_{n+1} = x_n \text{ to 6 decimal places}\\ \text{Use a spreadsheet or something}\\ \text{You should get }x = -0.254101 \)
Now just copy this method for problem 2 which has a slightly different recursion relation.