Given $\frac{\sqrt{2}\sin\alpha}{\sqrt{1+\cos 2\alpha}}=\frac17, \qquad \alpha\in\left(0,\frac\pi2\right)$ and $\sqrt{\frac{1-\cos 2\beta}{2}}=\frac{1}{\sqrt{10}}, \qquad \beta\in\left(0,\frac\pi2\right).$

Find $\alpha$

Use the identity $1+\cos 2\alpha = 2\cos^2\alpha.$ Since $\alpha\in\left(0,\frac\pi2\right)$, we have $\cos\alpha>0$, so $\sqrt{1+\cos 2\alpha}=\sqrt{2\cos^2\alpha}=\sqrt2\cos\alpha.$

Hence, $\frac{\sqrt2\sin\alpha}{\sqrt2\cos\alpha}=\tan\alpha=\frac17.$

So, $\tan\alpha=\frac17.$

Find $\beta$

Use the identity $1-\cos 2\beta = 2\sin^2\beta.$ Therefore, $\sqrt{\frac{1-\cos 2\beta}{2}}=\sqrt{\sin^2\beta}=\sin\beta$ because $\beta\in\left(0,\frac\pi2\right)$, so $\sin\beta>0$.

Thus, $\sin\beta=\frac{1}{\sqrt{10}}.$

Then

Find $\tan 2\beta$

Find $\tan(\alpha+2\beta)$

Using $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$ we get $\tan(\alpha+2\beta)=\frac{\tan\alpha+\tan 2\beta}{1-\tan\alpha\tan 2\beta}.$

Substitute $\tan\alpha=\frac17$ and $\tan 2\beta=\frac34$: $\tan(\alpha+2\beta)=\frac{\frac17+\frac34}{1-\frac17\cdot\frac34}.$

Numerator: $\frac17+\frac34=\frac{4+21}{28}=\frac{25}{28}.$

Denominator: $1-\frac{3}{28}=\frac{25}{28}.$

Therefore, $\tan(\alpha+2\beta)=\frac{25/28}{25/28}=1.$

Final Answer $\boxed{1}$

Comparison with stored answer

Stored correct answer = $1$.

My derived answer also equals $1$, so it agrees.