Answer
a) $\bf Potassium$
b) ${\bf 6.72\times 10^{-34}}\;\rm J\cdot s$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$V_{\rm stop}=\dfrac{h}{e}\left[ f-f_0\right]$$
where $f=c/\lambda$
$$V_{\rm stop}=\dfrac{h}{e}\left[ \dfrac{c}{\lambda}-f_0\right]\tag 1$$
So to find the metal, we need to find $E_0=hf_0$, and to find $E_0$, we need to find $f_0$.
So we can use equation (1) as a straight line equation $y=mx+b$;
where
$y=V_{\rm stop}$,
$m={\rm slope}=\dfrac{h}{e}$,
$x=f=\dfrac{c}{\lambda}$,
and it intersects $x$-axis at $x=f_0$
We are given the stop potential but we still need the frequencies.
$$\boxed{f =\dfrac{c}{\lambda}}$$
We will find these values and plug them as seen in the table below.
\begin{array}{|c|c|c|c|}
\hline
f \;(\times 10^{14}{\;\rm Hz})&V_{\rm stop} {\;\rm (V)}\\
\hline
6& 0.19 \\
\hline
6.67& 0.48 \\
\hline
7.5& 0.83 \\
\hline
8.57& 1.28 \\
\hline
10& 1.89 \\
\hline
12 & 2.74 \\
\hline
\end{array}
Now we just need to plug these given points into any software calculator to draw the best-fit line and hence we can find $f_0$.
It is obvious, from the graph below, that $f_0=\bf 5.55\times 10^{14}$ Hz, so
$$E_0=hf_0=(6.626\times 10^{-34})(5.55\times 10^{14})$$
$$E_0=\bf 3.677\times 10^{-19}\;\rm J=\bf 2.298\;\rm eV\approx \color{red}{\bf 2.3}\;\rm eV$$
So, according to Table 38.1, it is the $\underline{\color{magenta}{\bf potassium}}$.
$$\color{blue}{\bf [b]}$$
Using the slope formula which we explained above,
$$m={\rm slope}=\dfrac{h}{e}$$
From the figure below, we can see that the slope is given by
$${\rm slope}=\dfrac{h}{e}= \dfrac{0.42}{10^{14}} $$
Hence, the experimental value for Planck’s constant is given by
$$h= \dfrac{0.42e}{10^{14}}= \dfrac{0.42(1.6\times 10^{-19})}{10^{14}}$$
$$h=\color{red}{\bf 6.72\times 10^{-34}}\;\rm J\cdot s$$
