$$\log_{10}50 + \left(\frac{\log_{0.5}5}{1+\log_{2}5}\right)$$

$$\log_{0.5}5\ $$ can be written as $$\frac{\log_25}{\log_20.5}$$

It becomes -$$\log_25$$ as $$\log_20.5\ $$ is -1.

$$1+\log_25\ $$ will be $$\log_210\ $$ since 1 can be written as $$\log_22$$.

After these calculations $$ \left(\frac{\log_{0.5}5}{1+\log_{2}5}\right)$$ will trun into -$$\frac{\log_25}{\log_210}$$ which is -$$\log_{10}5$$

Now we have $$\log_{10}50 - \log_{10}5$$ which is nothing but $$\log_{10}10$$. So the answer is 1.