How to find $\sigma(\bar 9)$ if $\sigma(\bar5)=\bar6?$ 1)Let $\sigma:\Bbb Z_{12}\to\Bbb Z_{48}$ be a group homomorsiphism . find $\sigma(\bar 9)$ if $\sigma(\bar5)=\bar6$ ?
2)Let $\sigma:\Bbb Z_{10}\to\Bbb Z_{15}^*$ be a group homomorsiphism; $\bar a\in \Bbb Z_{10}$ and $\sigma(\bar a )=\bar 8$ find $\sigma(\bar {-a})$
at first question since $\sigma(\bar1)=a$ , I guess we need to solve $5a\equiv6 (mod\quad48)$ should I continue like $5a\equiv1 (mod\quad 48)$ and $48x+5y=1$ and solve this?
I have no idea about second one. can we say since it is a homo. $-\sigma(\bar a)=\sigma(\bar {-a})$

For the first one, I would instead solve $$5a=9\pmod{12},$$ so that $$\sigma\left(\overline 9\right)=a\sigma\left(\overline 5\right).$$ (Do you see why?)
For the second, bear in mind that $\Bbb Z_{15}^*$--the group of units of $\Bbb Z_{15}$-- is a group under multiplication, not addition. Hence, rather than $\sigma\left(-\overline a\right)=-\sigma\left(\overline a\right),$ we can say that $$\sigma\left(-\overline a\right)=\left(\sigma\left(\overline a\right)\right)^{-1}.$$
