You find it one number at a time. For (a), for example, find $a_1$ by plugging in $1$ everywhere you see an $n$. So, $a_1 = 7a_{1-1}-5 = 7a_0-5$. You are given that $a_0 = 1$, so $a_1 = 7(1)-5 = 2$. Then plug in $2$ for $n$: $a_2 = 7a_{2-1} - 5 = 7a_1-5 = 7(2) - 5 = 9$. Can you figure out $a_3, a_4, a_5$ from there?

Use the same process for (b) through (d).