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Dear Student, The solution/s to your 1st query has/have been provided below:

$$\begin{gathered} \left(a\right) let x=49.3\times 0.0842 \\ taking \log on both sides \\ \log x=\log (49.3\times 0.0842) \\ =\log 49.3 + \log 0.0842 \left\{\log \right(a\times b)=\log a + \log b\} \\ \log x=1.69+(-1.07) \\ \log x=0.61 \\ x={10}^{0.61}=4.07 \end{gathered}$$

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