$$ \text{Let for }f(x)=7 \tan^{8}x + 7\tan^{6}x-3\tan^{4}x-3\tan^{2}x \text{ } I_1=\int_{0}^{\pi/4}f(x)dx \text{ and }I_2=\int_{0}^{\pi/4}xf(x)dx. \text{ Then } 7I_1+12T_2 \text{ is equal to :} $$

We are given $$f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x$$ and need to find $$7I_1 + 12I_2$$ where $$I_1 = \int_0^{\pi/4} f(x)\,dx$$ and $$I_2 = \int_0^{\pi/4} x\,f(x)\,dx$$.

$$f(x) = 7\tan^6 x(\tan^2 x + 1) - 3\tan^2 x(\tan^2 x + 1)$$

$$= (\tan^2 x + 1)(7\tan^6 x - 3\tan^2 x)$$

$$= \sec^2 x \cdot \tan^2 x(7\tan^4 x - 3)$$

Substitute $$t = \tan x$$, so $$dt = \sec^2 x\,dx$$. When $$x = 0, t = 0$$; when $$x = \pi/4, t = 1$$.

$$I_1 = \int_0^1 t^2(7t^4 - 3)\,dt = \int_0^1 (7t^6 - 3t^2)\,dt = \left[t^7 - t^3\right]_0^1 = 1 - 1 = 0$$

From the substitution above: $$G(x) = \int f(x)\,dx = \tan^7 x - \tan^3 x$$

Let $$u = x$$ and $$dv = f(x)\,dx$$, so $$du = dx$$ and $$v = G(x) = \tan^7 x - \tan^3 x$$.

$$I_2 = \left[x \cdot G(x)\right]_0^{\pi/4} - \int_0^{\pi/4} G(x)\,dx$$

$$= \frac{\pi}{4}(\tan^7(\pi/4) - \tan^3(\pi/4)) - \int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx$$

$$= \frac{\pi}{4}(1 - 1) - \int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx$$

$$= -\int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx$$

Using the reduction formula $$\int_0^{\pi/4} \tan^n x\,dx = \frac{1}{n-1} - \int_0^{\pi/4} \tan^{n-2} x\,dx$$:

$$\int_0^{\pi/4} \tan^3 x\,dx = \frac{1}{2} - \int_0^{\pi/4} \tan x\,dx = \frac{1}{2} - \frac{\ln 2}{2}$$

$$\int_0^{\pi/4} \tan^5 x\,dx = \frac{1}{4} - \left(\frac{1}{2} - \frac{\ln 2}{2}\right) = -\frac{1}{4} + \frac{\ln 2}{2}$$

$$\int_0^{\pi/4} \tan^7 x\,dx = \frac{1}{6} - \left(-\frac{1}{4} + \frac{\ln 2}{2}\right) = \frac{5}{12} - \frac{\ln 2}{2}$$

Therefore:

$$\int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx = \left(\frac{5}{12} - \frac{\ln 2}{2}\right) - \left(\frac{1}{2} - \frac{\ln 2}{2}\right) = -\frac{1}{12}$$

$$I_2 = -\left(-\frac{1}{12}\right) = \frac{1}{12}$$

$$7I_1 + 12I_2 = 7(0) + 12 \cdot \frac{1}{12} = 0 + 1 = 1$$

The answer is Option B: 1.