For a start, neither (1) nor (2) are given to you as premisses. Your premiss is the given conditional, and then your proof should proceed by assuming $B$ for reductio ...

To be honest, your proof attempt suggests you really haven't understood ND at all. For a gentle and I hope very accessible intro to ND (which is quite freely available) try the relevant chapters of my book Introduction to Formal Logic which you can download from

Natural Deduction's rules of inference revolve around 'introducing' or 'eliminating' logical connectives.   So that is where we start.$\def\fitch#1#2{~~~~~\begin{array}{|l}#1\\\hline#2\end{array}}$

You are given a premise $B\to(A\land\lnot A)$ and wish to infer $\lnot B$.   Well there is a conditional and conjunction to eliminate in the premise, and a negation to introduce in the conclusion. Clearly these rules need to be understood (at least).

$$\dfrac{\phi\to\chi\quad\phi}{\chi}{~\small{\to}\mathsf e}\qquad\dfrac{\gamma\wedge\beta}{\gamma}{~\small{\land}\mathsf e}\qquad\dfrac{\gamma\wedge\beta}{\beta}{~\small{\land}\mathsf e}\qquad\dfrac{[\psi]\\~\vdots\\\bot}{\lnot\psi~} {~\small\lnot\mathsf i}$$

So... to successfully apply negation introduction, an assumption must be made, under which a contradiction must be derived.   For this proof that assumption must be $B$, and this and the premise then allows conditional elimination to be applied, thus deriving $A\land\lnot A$.   Then from this... well...

$$\fitch{~~1.~~B\to(A\land\lnot A)\hspace{7ex}\textsf{Premise}}{\fitch{~~2.~~B\hspace{16ex}\textsf{Assumption}}{~~3.~~A\land\lnot A\hspace{10ex}{\to}\mathsf e~1,2\\~~4.~~A\hspace{16ex}{\land}\mathsf e~3\\~~5.~~\lnot A\hspace{14.5ex}{\land}\mathsf e~3\\~~~~\vdots\\~~\mathrm m.~~\bot}\\~~\mathrm n.~~\lnot B\hspace{17ex}\lnot\mathsf i~2{-}\mathrm m}$$

So all that is left is to derive that contradiction.