Prove $\log(n^2 + 1)$ is $Θ(\log n)$
In order to prove that $\log(n^2+1) \in \Theta(\log n)$ according to your definition, you have to produce $a,b,n_0$ such that $a \log n \leq \log(n^2+1) \leq b\log n$ for all $n \geq n_0$. This is not what you did. What you have shown is some other inequality.

To help you in future such questions, here is a complete proof. We choose $a = 2$, $b = 3$, and $n_0 = 2$. So we have to show that for $n \geq 2$,
$$
2\log n \leq \log(n^2+1) \leq 3\log n.
$$

The first inequality follows from
$$
\log(n^2+1) \geq \log(n^2) = 2\log n.
$$
You did that part correctly.

The second inequality follows from
$$
\log(n^2+1) \leq \log(n^2+n^2) = \log(2n^2) \leq \log(n^3) = 3\log n.
$$
This is the part you did incorrectly.