1. modular proof

Prove that if a=bmodn and c=dmodn then a+c=b+dmodn and ac=bdmodn
My solution for the first part is:

n|(a-b) so (a-b)=nq and n|(c-d) so (c-d)=np
so (a-b) +(c-d)=nq+np
(a+c)-(b+d)=n(q+p)
So n|[(a+c)-(b+d)]
So
a+c=b+dmodn
I am not sure how to show the second part any help would be great

2. Same kind of proof works.

$a \equiv b \ (\text{mod } n) \ \Leftrightarrow \ a - b = nq$
$c \equiv d \ (\text{mod } n) \ \Leftrightarrow \ c - d = np$

So:
\begin{aligned} ac - bd & = ac - (a-nq)(c - np) \\ & = \cdots \\ & = n \left(anp + cnq - npq\right) \end{aligned}

which by definition, is what we're looking for.