Using Diophantine equation

If $x\equiv16\bmod78$ and $x\equiv15\bmod77$, then $x\equiv0\bmod2$ and $x\equiv1\bmod 7$.

Can you take it from here?

But if you insist on doing it your way,

$a=77x+15=78y+16$ with $x=y=-1$

so $a\equiv 77(78n-1)+15\equiv 78(77m-1)+16\bmod (77\times78)\equiv5944\bmod6006$.

Now can you find the remainder mod $14$?

For $x_0=-1$, $a=77(-1)+15=-62$ is indeed one of the infinitely many integers which are solutions to the question.

So just find $-62 \pmod {14}$.

$$N = 77x + 15=78y + 16$$

$$N+62 = 77x + 77=78y + 78$$

$$LCM(78;77)=6006$$

$$N+62=6006k, k \in Z$$

$$6006k-62\equiv -62\equiv 8 \pmod {14}$$