Frequency $f$ is given by the period $T$. The period can be measured in the imagine of the scope.

1. The sine waves repeat after $6 ~\rm divisions$ (e.g. from falling turning point to falling turning point of one curve)
2. The scale is $0.1 ~\rm ms/Div$

\begin{align*} f &= {{1} \over {T}} \\ T &= 6 ~\rm Div \cdot 0.1 ~ms/Div \\ \rightarrow f &= {{1} \over {6 ~\rm Div \cdot 0.1 ~ms/Div}} \end{align*}

\begin{align*} f = 1.66... ~{\rm kHz} \rightarrow f &= 1.67 ~\rm kHz \end{align*}

Similar to 1. the amplitude can be derived. The amplitude is given by the voltage difference between maximum and turning point.

1. The amplitude of the voltage is $3 ~\rm divisions$
2. The scale is $5 ~\rm V/Div$

\begin{align*} \hat{U} &= 3 ~\rm Div \cdot 5 ~\rm V/Div \end{align*}

Similar for the current.

\begin{align*} \hat{U} &= 15 ~\rm V \\ \hat{I} &= 0.5 ~\rm A \end{align*}

The RMS values for sine waves are given as \begin{align*} \square_{\rm RMS} ={{1} \over {2} } \sqrt{2} \cdot \hat{\square} \end{align*}

\begin{align*} U_{\rm RMS} & ={{1} \over {2} } \sqrt{2} \cdot \hat{U} ={{1} \over {2} } \sqrt{2} \cdot 15 ~\rm V \\ I_{\rm RMS} & ={{1} \over {2} } \sqrt{2} \cdot \hat{I} ={{1} \over {2} } \sqrt{2} \cdot 0.5 ~\rm A \end{align*}

\begin{align*} U_{\rm RMS} & = 10.606... ~\rm V \rightarrow U_{\rm RMS} = 10.6 ~\rm V \\ I_{\rm RMS} &= 0.3535... ~\rm A \rightarrow I_{\rm RMS} = 0.35 ~\rm A \end{align*}

The phase shift $\varphi$ is given as the angle between the current and voltage phasor. The phase shift is negative when the voltage lags the current.
A full period has an angle of $360°$ or $2\pi$ in $\rm radian$.
One has to find out the phase shift as a fraction of a period to get the value of the phase shift:

\begin{align*} \varphi &= \rm -{{0.5 ~Div} \over { 6 ~Div}} \cdot 2\pi \end{align*}

\begin{align*} \varphi &= -{{1} \over { 6}} \pi \\ \varphi &= - 30° \end{align*}