If $$x_{1}, x_{2}$$ are the roots of the equation, $$2x^{2} — 4x + 5 = 0$$, then the equation whose roots are $$x_{1} + \frac{1}{x_{1}}$$ and $$x_{2} + \frac{1}{x_{2}}$$ is:

After seeing that the initial quadratic equation cannot be easily factorised, we should be looking at the sum and product of the roots, 
Sum of the roots: $$x_1+x_2=2$$
Product of roots: $$x_1x_2=\frac{5}{2}$$

Now for the new equation we need, we can again calculate the sum and product of roots. 
Sum of roots: $$x_1\ +x_2\ +\frac{1}{x1}+\frac{1}{x2}$$
                     =$$\left(x_1+x_2\right)+\left(\frac{x_1+x_2}{x_1x_2}\right)$$
                     =$$\left(x_1+x_2\right)\left(1+\frac{1}{x_1x_2}\right)$$  
Putting in the values, we get: $$2\left(1+\frac{2}{5}\right)=\frac{14}{5}$$

Product of roots: $$\left(x_1+\frac{1}{x_1}\right)\left(x_2+\frac{1}{x_2}\right)$$
                          =$$x_1x_2+\frac{x_1}{x_2}+\frac{x_2}{x_1}+\frac{1}{x_1x_2}$$
                          =$$x_1x_2+\frac{1}{x_1x_2}+\frac{\left(x_1^2+x_2^2\right)}{x_1x_2}$$
                          =$$x_1x_2+\frac{1}{x_1x_2}+\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}$$
Putting in the values, we get $$\frac{5}{2}+\frac{2}{5}+\frac{2\left(\left(2\right)^2-\left(2\times\ \frac{5}{2}\right)\right)}{5}$$
                          =$$\frac{5}{2}+\frac{2}{5}-\frac{2}{5}$$
                          =$$\frac{5}{2}$$

Checking the options for these conditions of sum and product of roots, we see that only option A satisfies all of our criterion.
Therefore, Option A is the correct answer.