$$ \\ 0 \to A \to B \to C \to 0 \\ 0 \to A' \to B' \to C' \to 0 $$ These are exact and they occur a commutative diagram by homomorphism. $$ g=B\to B'\\ f=A\to A'\\ h=C \to C' $$ Suppose that $g$ is isomorphism. Then $f$ is monic and $h$ is epic.
To find $f$ is monic I need to find $\ker(f)=\{0\}$. Suppose $a$ is in $\ker(f)$. $$( x= A \to B , x'=A' \to B' , y=B \to C , y'= B' \to C' )$$ Also we have $g(b)=y'(a')$. By diagram, $x'f(a)=gx(a)$. $f(a)=0$ ( $a$ is in $\ker f$) then $x'(0)=0$. because $x'$ is homomophism. Then $g(x(a))=0$. it means that $x(a)$ is in $\ker(g)$.
How can I get $a =0$?
if $h$ is epic,then there is $c'$ such that $h(c)=c'$. By diagram , $y'x'(a')=hy(b)=0$.
how can I find $c'$? Can someone help me out, please ?
You were close concerning $f$! Let me help you out.
Proof. Let $x\in\ker f$. Then by commutativity $(g\circ\alpha_0)(x)=(\alpha_1\circ f)(x)=0$ and therefore $\alpha_0(x)\in\ker g$. But $g$ is an isomorphism, thus $\ker g=\{0\}$ and so $\alpha_0(x)=0$. As $\alpha_0$ is monic it immediately follows that $x=0$ as desired.
Now, take $x\in C'$. Then, as $\beta_1$ is epic and $g$ is an isomorphism, $\beta_1\circ g$ is epic aswell. So there is $y\in B$ such that $(\beta_1\circ g)(y)=x$. Now, let $z=\beta_0(x)\in C$. By commutativity $$h(z)=(h\circ\beta_0)(y)=(\beta_1\circ g)(y)=x$$ Thus $h$ is epic as desired. $\square$
Proof. Take $x\in A'$ and consider $\alpha_1(x)$. Then there is $y\in B$ such that $g(y)=\alpha_1(x)$ as $g$ is an isomorphism. By commutativity $$(h\circ\beta_0)(y)=(\beta_1\circ g)(y)=(\beta_1\circ\alpha_1)(x)=0$$ Thus $\beta_0(y)\in\ker h$. But $h$ is monic by assumption and therefore $\beta_0(y)=0$ and $y\in\ker\beta_0$. By exactness there is $z\in A$ such that $\alpha_0(z)=y$. Then it is immediate that $$(\alpha_1\circ f)(z)=(g\circ\alpha_0)(z)=g(y)=\alpha_1(x)$$ As $\alpha_1$ is monic we obtain $f(z)=x$ as desired. $\square$