1. Let

$$\theta=2\tan^{-1}\frac15+\sec^{-1}\frac{\sqrt5}{2}+2\tan^{-1}\frac18.$$

We need to find $\tan\theta$.

2. First, simplify each part.

(i) Compute $\tan\left(2\tan^{-1}\frac15\right)$

Use

$$\tan(2A)=\frac{2\tan A}{1-\tan^2 A}.$$

Here $\tan A=\frac15$, so

$$\tan(2A)=\frac{2\cdot \frac15}{1-\left(\frac15\right)^2} =\frac{\frac25}{1-\frac1{25}} =\frac{\frac25}{\frac{24}{25}} =\frac{5}{12}.$$

So,

$$\tan\left(2\tan^{-1}\frac15\right)=\frac{5}{12}.$$

(ii) Compute $\tan\left(\sec^{-1}\frac{\sqrt5}{2}\right)$

Let

$$B=\sec^{-1}\frac{\sqrt5}{2}.$$

Then

$$\sec B=\frac{\sqrt5}{2}.$$

But for real values of $\sec^{-1}x$, we must have

$$|x|\ge 1.$$

Here

$$\frac{\sqrt5}{2}\approx 1.118>1,$$

so this is valid. Then

$$\cos B=\frac{2}{\sqrt5}.$$

Since principal value of $\sec^{-1}$ lies in $[0,\pi]$, excluding $\pi/2$, and $\cos B>0$, we get $B$ in first quadrant. Hence

$$\sin B=\sqrt{1-\cos^2 B}= \sqrt{1-\frac{4}{5}}=\frac{1}{\sqrt5}.$$

Therefore,

$$\tan B=\frac{\sin B}{\cos B} =\frac{\frac1{\sqrt5}}{\frac2{\sqrt5}}=\frac12.$$

So,

$$\tan\left(\sec^{-1}\frac{\sqrt5}{2}\right)=\frac12.$$

(iii) Compute $\tan\left(2\tan^{-1}\frac18\right)$

Again using

$$\tan(2C)=\frac{2\tan C}{1-\tan^2 C},$$

with $\tan C=\frac18$,

$$\tan(2C)=\frac{2\cdot \frac18}{1-\left(\frac18\right)^2} =\frac{\frac14}{1-\frac1{64}} =\frac{\frac14}{\frac{63}{64}} =\frac{16}{63}.$$

So,

$$\tan\left(2\tan^{-1}\frac18\right)=\frac{16}{63}.$$

3. Now let

$$X=2\tan^{-1}\frac15,\quad Y=\sec^{-1}\frac{\sqrt5}{2},\quad Z=2\tan^{-1}\frac18.$$

Then

$$\tan X=\frac{5}{12},\quad \tan Y=\frac12,\quad \tan Z=\frac{16}{63}.$$

We need

$$\tan(X+Y+Z).$$

First compute $\tan(X+Y)$:

$$\tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y} =\frac{\frac{5}{12}+\frac12}{1-\frac{5}{12}\cdot\frac12}.$$

Now,

$$\frac{5}{12}+\frac12=\frac{5}{12}+\frac{6}{12}=\frac{11}{12},$$

and

$$1-\frac{5}{24}=\frac{19}{24}.$$

Hence,

$$\tan(X+Y)=\frac{\frac{11}{12}}{\frac{19}{24}}=\frac{11}{12}\cdot\frac{24}{19}=\frac{22}{19}.$$

4. Now compute $\tan(X+Y+Z)$:

$$\tan(X+Y+Z)=\frac{\tan(X+Y)+\tan Z}{1-\tan(X+Y)\tan Z} =\frac{\frac{22}{19}+\frac{16}{63}}{1-\frac{22}{19}\cdot\frac{16}{63}}.$$

Take LCM in numerator:

$$\frac{22}{19}+\frac{16}{63} =\frac{22\cdot 63+16\cdot 19}{19\cdot 63} =\frac{1386+304}{1197} =\frac{1690}{1197}.$$

Denominator:

$$1-\frac{22\cdot 16}{19\cdot 63} =1-\frac{352}{1197} =\frac{1197-352}{1197} =\frac{845}{1197}.$$

Thus,

$$\tan(X+Y+Z)=\frac{\frac{1690}{1197}}{\frac{845}{1197}}=\frac{1690}{845}=2.$$

5. Therefore,

$$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)=2.$$

So the correct option is B.