Proofs with inequalities I am asked to prove the following, given that $a<b$, and $c<d$, such that $a,b,c,d>0$:

(1) $a+c<b+d$
(2) $ac<bd$

I have been struggling with these kinds of questions in my course, and am also wondering if there's a good way to approach these kinds of questions.

We want to show the following two things, (1) $a+c<b+d$, and (2) $ac<bd$.
The first is quite straightforward. First, we add $c$ to $a<b\to a+c<b+c$. Then, add, $b$ to both sides of $c<d\to b+c<b+d$. Hence, we have $$a+c<b+c<b+d\to a+c<b+d,$$ which is what we wanted to show.
For the second part, multiply $a<b$ by $c$. (Granted that $a,b,c,d>0$, we won't worry about reversal of the inequality when multiplying through by them.) We have, $ac<bc$. Then, multiply $c<d$, by $b$. We have $bc<bd$, and so, $$ac<bc<bd\to ac<bd,$$ which is again, what we wanted to show.
It is useful to have a list of fundamental rules next to you when you are approaching questions like this.
