If you have a short exact sequence
$0\rightarrow\mathscr{F}\rightarrow\mathscr{G}\rightarrow\mathscr{H}\rightarrow 0$
of finite locally free sheaves on a scheme $X$, then the sequence
$0\rightarrow\mathscr{H}^\vee\rightarrow\mathscr{G}^\vee\rightarrow\mathscr{F}^\vee\rightarrow 0$
is exact. The reason is that exactness can be checked on stalks, and because the sheaves in the original sequence are finitely presented, taking stalks commutes with taking $\mathcal{H}om$ sheaves, so the sequence of stalks of the second sequence is
$0\rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{H}_x,\mathscr{O}_{X,x}) \rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{G}_x,\mathscr{O}_{X,x}) \rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{O}_{X,x})\rightarrow 0$
which is exact because the functor $\mathrm{Hom}_{\mathscr{O}_{X,x}}(-,\mathscr{O}_{X,x})$ is exact on short exact sequences of finite free $\mathscr{O}_{X,x}$-modules.
I'm not sure what happens when the sheaves in the sequence are not finite locally free. There is a long exact sequence of $\mathcal{E}xt$ sheaves. Note that the sheaves in the OP's original sequence are all finite locally free.
EDIT: Incidentally, this has nothing to do with schemes, and works for arbitrary locally ringed spaces.