(i) Using the identity \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute \(\tan x = t\) into the equation:

\(2 \left(\frac{2t}{1-t^2}\right) + 5t^2 = 0\)

Simplifying gives:

\(4t + 5t^2 - 5t^4 = 0\)

Factoring out \(t\), we get:

\(t(4 + 5t - 5t^3) = 0\)

Thus, \(t = 0\) or \(t = \sqrt[3]{(t + 0.8)}\).

(ii) Check the sign of \(t - \sqrt[3]{(t + 0.8)}\) at \(t = 1.2\) and \(t = 1.3\):

At \(t = 1.2\), \(1.2 - \sqrt[3]{(1.2 + 0.8)} = -0.06\)

At \(t = 1.3\), \(1.3 - \sqrt[3]{(1.3 + 0.8)} = 0.02\)

There is a change of sign, confirming a root between 1.2 and 1.3.

(iii) Using the iterative formula \(t_{n+1} = \sqrt[3]{(t_n + 0.8)}\):

Start with \(t_0 = 1.2\):

\(t_1 = \sqrt[3]{(1.2 + 0.8)} = 1.2765\)

\(t_2 = \sqrt[3]{(1.2765 + 0.8)} = 1.2760\)

\(t_3 = \sqrt[3]{(1.2760 + 0.8)} = 1.2760\)

The value of \(t\) correct to 3 decimal places is 1.276.

(iv) Solve \(2 \tan 2x + 5 \tan^2 x = 0\) using \(t = 1.276\):

\(\tan x = 1.276\), so \(x = \arctan(1.276)\)

\(x \approx 0.906\)

Considering the periodicity of \(\tan x\), the solutions are:

\(x = -2.24, 0.906, \pi\)