For 6.7, let's look at a smaller example to see if we can find a pattern.

Suppose we have the same recursive definition, except now we want to just find $a_4$. This is small enough that we can write out the definition (even if it is messy): $$a_4 = 3^{a_3} = 3^{\displaystyle 3^{a_2}} = 3^{\displaystyle 3^{\displaystyle 3^{a_1}}} = 3^{\displaystyle 3^{\displaystyle 3^{1}}}$$ Now, let's ignore the fact that these numbers are small enough to calculate directly to try to see how we could approach the problem in general.

Using patterns or Fermat's Little Theorem, we know that $3^6\equiv 1 \pmod{7}$. Therefore, we know that only the remainder when we divide by $6$ matters for the exponent when we are working mod 7. We can write: $$3^{\displaystyle 3^{\displaystyle 3^{1}}} \equiv \displaystyle 3^{\displaystyle 3^{\displaystyle 3^{1}}\pmod{6}} \pmod{7}.$$ So we need to find the remainder when $3^{\displaystyle 3^{1}}$ is divided by $6$. Is there a pattern in the powers of $3$ working mod 6?