Solution:

$(1)$ From the given conditions, we have: $\begin{cases}7a_1+ \frac{7×6}{2}d=35 \\ \left(a_1+4d\right)^2=\left(a_1+d\right)\left(a_1+10d\right)\end{cases}$,

which simplifies to $\begin{cases}a_1+3d=5 \\ 2d=a_1\end{cases}$,

Since $d \neq 0$,

we solve to get $\begin{cases}a_1=2 \\ d=1\end{cases}$,
Therefore, $a_n=2+(n-1)=n+1$.

$(2)$ $\frac{1}{a_na_{n+1}}= \frac{1}{(n+1)(n+2)}= \frac{1}{n+1}- \frac{1}{n+2}$,
Therefore, $T_n=\left( \frac{1}{2}- \frac{1}{3}\right)+\left( \frac{1}{3}- \frac{1}{4}\right)+...+\left( \frac{1}{n+1}- \frac{1}{n+2}\right)= \frac{1}{2}- \frac{1}{n+2}$,
The inequality $T_n-\lambda a_{n+1} \geqslant 0$, becomes $\frac{1}{2}- \frac{1}{n+2}-\lambda(n+2) \geqslant 0$,

which simplifies to: $\lambda \leqslant \frac{n}{2(n+2)^2}$,
Since $\frac{n}{2(n+2)^2}= \frac{1}{2(n+ \frac{4}{n}+4)}\leqslant \frac{1}{2×(2 \sqrt{4}+4)}= \frac{1}{16}$,

equality holds only when $n=2$,
Since there exists $n \in \mathbb{N}^*$ such that $T_n-\lambda a_{n+1} \geqslant 0$ holds,
Therefore, the range of values for the real number $\lambda$ is $\boxed{(-\infty, \frac{1}{16}]}$.