Consider the given graph showing variation of reactant...

Analysis of the graph:
The graph shows reactant concentration $[R]$ vs time $t$ for three reactions starting from the same initial concentration.

Curve 1 is a straight line with negative slope. For a zero-order reaction, the integrated rate law is $[R] = [R]_0 - kt$, which is a straight line in $[R]$ vs $t$. Hence, reaction $1$ is a zero-order reaction.

Curves 2 and 3 are non-linear (exponential-type decays), so they correspond to reactions of order greater than zero (e.g., first or second order).

Evaluating the options:

$(1)$ The order of all three reactions is the same.
Reaction $1$ is zero order while reactions $2$ and $3$ are of higher order, so this statement is incorrect.

$(2)$ The rate constant of reaction $3$ is larger than that of reaction $2$, assuming both have the same order.
Assuming both curves $2$ and $3$ correspond to the same order (say first order), the integrated rate law is $[R] = [R]_0 e^{-kt}$. At any given time $t$, the curve for reaction $3$ lies below that of reaction $2$, i.e., $[R]_3 < [R]_2$. Since both start from the same $[R]_0$, reaction $3$ is decaying faster, which implies a larger rate constant. Hence $k_3 > k_2$, and this statement is correct.

$(3)$ The SI unit of the rate constant of reaction $1$ is $\text{s}^{-1}$.
For a zero-order reaction, $\text{Rate} = k[R]^0 = k$, so the unit of $k$ is that of rate, i.e., $\text{mol L}^{-1}\text{s}^{-1}$. The unit $\text{s}^{-1}$ corresponds to a first-order reaction, so this statement is incorrect.

$(4)$ Thermal decomposition of $HI$ on a gold surface is an example of reaction $2$.
The thermal decomposition of $HI$ on a gold surface is a well-known zero-order reaction, which corresponds to curve $1$, not curve $2$. Hence this statement is incorrect.

Therefore, the correct option is $(2)$.