Q:

Prove if $a\in A$ is a maximum then $f(a)\in B$ is maximum and if $(A,\le_A)$ is totally ordered then $(B,\le_B)$ is totally ordered

Two ordered sets $(A,\le_A), (B,\le_B)$ and there's an isomorphic function $f:A\to B$

Prove if $a\in A$ is a maximum then $f(a)\in B$ is maximum.

if $(A,\le_A)$ is totally ordered then  $(B,\le_B)$ is totally ordered.

Suppose by contrapostion $f(a)\in B$ isn't the maximum, then either $B$ doesn't have a maximum and that is a contradiction because there can't be isomorphism between one set that has a maximum and another that doesn't have one. Or $max \{B\}\neq f(a)$ which is also a contradiction to the order preserving of isomorphism.

So if one set is totally ordered that means $\forall a,b\in A$ we have $a\le b$, applying the isomorphic function we get $f(a)\le f(b): \forall f(a),f(b)\in B$, we know from isomorphism that both sets have the same cardinality so the image of $f$ must be equal to $B$ and from $f(a)\le f(b): \forall f(a),f(b)\in B$ we'll get that  $B$ is also totally ordered.

A:

Your arguments are a bit sloppy IMHO. 
(1) Let $b \in B$ be arbitrary. Pick $a' \in A$ such that $f(a') = b$. This can be done as $f$ is a bijection, so a surjection in particular. Then $a' \le_A a$ as $a$ is a maximum. So $b = f(a') \le_B f(a)$. As $b$ was arbitrary, $f(a)$ is a maximum in $B$.
No contraposition needed.
(2). Pick any two distinct $b_1, b_2$ in $B$. Again by surjectivity, we find $a_1, a_2 \in A$ such that $f(a_1) = b_1, f(a_2) = b_2$. As $A$ is totally ordered we know that either $a_1 \le_A a_2$ or $a_2 \le_A a_1$. But then (in the first case) also $b_1 = f(a_1) \le_B f(a_2) = b_2$, and the second case is the same. So $B$ is totally ordered by $\le_B$.

