Step-by-step Solution:

The given function is $f(x) = 7\tan^8x + 7\tan^6x - 3\tan^4x - 3\tan^2x$ for all $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right).$ We need to evaluate the integrals given in the options.

Step 1: Simplify the function f(x)

We can simplify the expression for $f(x)$ by factoring out common terms: $f(x) = 7\tan^6x(\tan^2x + 1) - 3\tan^2x(\tan^2x + 1)$ Using the trigonometric identity $1 + \tan^2x = \sec^2x$, we get: $f(x) = 7\tan^6x \sec^2x - 3\tan^2x \sec^2x$

Step 2: Evaluate $\int_0^{\pi/4} f(x) dx$ (for options B and D)

Let's evaluate the definite integral of $f(x)$ from $0$ to $\pi/4$: $I_1 = \int_0^{\pi/4} f(x) dx = \int_0^{\pi/4} (7\tan^6x \sec^2x - 3\tan^2x \sec^2x) dx$ This integral is easy to solve using a substitution. Let $t = \tan x$. Then $dt = \sec^2x dx$. We also need to change the limits of integration:

• When $x = 0$, $t = \tan(0) = 0$.
• When $x = \pi/4$, $t = \tan(\pi/4) = 1$.

The integral becomes: $I_1 = \int_0^1 (7t^6 - 3t^2) dt$ Now, we integrate with respect to $t$: $I_1 = \left[ 7 \frac{t^7}{7} - 3 \frac{t^3}{3} \right]_0^1 = \left[ t^7 - t^3 \right]_0^1$ $I_1 = (1^7 - 1^3) - (0^7 - 0^3) = (1 - 1) - 0 = 0$ So, $\int_0^{\pi/4} f(x) dx = 0$. This confirms that option B is correct and option D is incorrect.

Step 3: Evaluate $\int_0^{\pi/4} xf(x) dx$ (for options A and C)

Let's evaluate the integral: $I_2 = \int_0^{\pi/4} xf(x) dx = \int_0^{\pi/4} x(7\tan^6x \sec^2x - 3\tan^2x \sec^2x) dx$ We will use integration by parts, with the formula $\int u \, dv = uv - \int v \, du$. Let $u = x$ and $dv = f(x) dx = (7\tan^6x \sec^2x - 3\tan^2x \sec^2x) dx$. Then $du = dx$. To find $v$, we integrate $dv$: $v = \int (7\tan^6x \sec^2x - 3\tan^2x \sec^2x) dx$ Using the same substitution as in Step 2 ($t=\tan x$), we found that this integral is $t^7 - t^3$. So, $v = \tan^7x - \tan^3x$

Now, applying the integration by parts formula: $I_2 = \left[ x(\tan^7x - \tan^3x) \right]_0^{\pi/4} - \int_0^{\pi/4} (\tan^7x - \tan^3x) dx$

First, evaluate the bracketed term: $\left[ x(\tan^7x - \tan^3x) \right]_0^{\pi/4} = \frac{\pi}{4}(\tan^7(\frac{\pi}{4}) - \tan^3(\frac{\pi}{4})) - 0(\tan^7(0) - \tan^3(0))$ $= \frac{\pi}{4}(1^7 - 1^3) - 0 = \frac{\pi}{4}(1 - 1) = 0$

So, the integral simplifies to: $I_2 = - \int_0^{\pi/4} (\tan^7x - \tan^3x) dx = \int_0^{\pi/4} (\tan^3x - \tan^7x) dx$ $I_2 = \int_0^{\pi/4} \tan^3x(1 - \tan^4x) dx$ We can factor $1 - \tan^4x = (1 - \tan^2x)(1 + \tan^2x) = (1 - \tan^2x)\sec^2x$. $I_2 = \int_0^{\pi/4} \tan^3x(1 - \tan^2x)\sec^2x dx$

Again, let $t = \tan x$, so $dt = \sec^2x dx$. The limits are from 0 to 1. $I_2 = \int_0^1 t^3(1 - t^2) dt = \int_0^1 (t^3 - t^5) dt$ $I_2 = \left[ \frac{t^4}{4} - \frac{t^6}{6} \right]_0^1$ $I_2 = \left( \frac{1^4}{4} - \frac{1^6}{6} \right) - (0 - 0) = \frac{1}{4} - \frac{1}{6}$ $I_2 = \frac{3 - 2}{12} = \frac{1}{12}$ So, $\int_0^{\pi/4} xf(x) dx = \frac{1}{12}$. This confirms that option A is correct and option C is incorrect.

Conclusion: Based on our calculations, the correct expressions are given in options A and B.