Proof that $\dim(U \times V) = \dim U + \dim V$. The following theorem in Serge Lang's Linear Algebra is left as an exercise, namely,
Let $U$ and $V$ be finite dimensional vector spaces over a field $K$, where $\dim U = n$ and $\dim V = m$. Then  $\dim W = \dim U + \dim V$, where $W = U \times V$, the direct product of the two vector spaces $U$ and $V$. Namely, $W$ contains the set of all ordered pairs $(u,w)$ such that $u \in U$ and $v \in V$. The usual axioms for such a direct product are:
1) Addition is defined component wise, namely if $(u_1,v_1),(u_2,v_2) \in W$, then $(u_1,v_1)+(u_2,v_2) = (u_1 + u_2, v_1 + v_2)$;
2)If $c \in K$, then $c(u_1,w_1) = (cu_1,cw_1)$.
To prove it, let $(u_1, u_2 \ldots u_n)$ be a basis for $U$ and $(v_1,v_2 , \ldots v_m)$ a basis for $V$.
So by definition, every element of $W$ can be written in the form 
$(a_1u_1 + \ldots a_nu_n, b_1v_1 + \ldots b_mv_m)$, where the $a_i's$ and $b_j's$ belong to the field $K$.
Using the above axioms this can be rewritten as:
$a_1(u_1,0) + a_2(u_2,0) + \ldots a_n(u_n,0) + b_1(0,v_1) + \ldots b_m(0,v_m)$.
Doubt: If we view all the $(u_i,0)$ ordered pairs as being the "basis" vectors of $U$ and similarly for the $(0,v_j)$ ordered pairs of $V$, then there are $n+m$ number of them and so proving the linear independence of these objects should suffice. But I'm confused because I know that $u_i's$ by themselves are the basis vectors of $U$, but now we are talking about ordered pairs $(u_i,0)$. How can I get out of such a situation?
Perhaps one can define some linear map between say a $u_i$ and the ordered pair $(u_i,0)$.
$\textbf{Edit}:$ First it is easy to see that $(U \times \{0\}) \cap (\{0\} \times V)$ is the ordered pair $(0,0)$. The linear independence of the basis vectors as stated above then follows.

Proof of exercise 1: Say that the $v_i$ are linearly independent. Then
$c_1v_1 + c_2v_2 + \ldots c_nv_n = 0$ only when all the $c_i = 0$. Applying $T$ to both sides we have 
$c_1T(v_1) + c_2T(v_2) + \ldots c_nT(v_n) = 0$ by linearity of $T$ and the fact that it is injective.
But the assumption was that all the $c_i$ were zero in order for the first equation to hold.
It follows that $(T(v_1), T(v_2), \ldots T(v_n))$ are a linearly independent set of vectors.
As for exercise 2, is it not clear from what it means for two ordered pairs to be equal that $f$ and $g$ are injective?
