Proof of (1):

Step 1: For $n=1$, we start by substituting $n=1$ into the given equation $\frac{2}{3}S_{n}=a_{n}-\frac{2}{3}n-2$ to find $a_1$:
$$\frac{2}{3}S_{1}=a_{1}-\frac{2}{3}\cdot1-2 \implies a_{1}=8.$$

Step 2: For $n\geqslant 2$, we consider the sum of the first $n-1$ terms:
$$\frac{2}{3}S_{n-1}=a_{n-1}-\frac{2}{3}(n-1)-2.$$

Step 3: Subtracting the equation for $S_{n-1}$ from the equation for $S_n$ gives us:
$$\frac{2}{3}S_{n}-\frac{2}{3}S_{n-1}=a_{n}-a_{n-1}-\frac{2}{3} \implies 2a_{n}=3a_{n-1}+2.$$

Step 4: Solving for $a_n$ in terms of $a_{n-1}$, we find:
$$a_{n}=3a_{n-1}+2 \implies a_{n}+1=3(a_{n-1}+1).$$

Therefore, the sequence $\{a_{n}+1\}$ is a geometric sequence with the first term $a_{1}+1=9$ and common ratio $3$. Hence, we have $\boxed{\text{The sequence } \{a_{n}+1\} \text{ is a geometric sequence.}}$

Proof of (2):

Step 1: From (1), we know $a_{n}+1=9\times 3^{n-1}=3^{n+1}$. Therefore, $a_{n}=3^{n+1}-1$.

Step 2: For the sequence $\{b_{n}\}$, we have:
$$b_{n}=\frac{1}{a_{n}+2}=\frac{1}{{3}^{n+1}+1} \lt \frac{1}{{3}^{n+1}}.$$

Step 3: The sum of the first $n$ terms of $\{b_{n}\}$, $T_{n}$, can be estimated as:
$$T_{n} < \frac{1}{{3}^{2}}+\frac{1}{{3}^{3}}+\frac{1}{{3}^{4}}+\cdots +\frac{1}{{3}^{n+1}}.$$

Step 4: Recognizing this as a geometric series with the first term $\frac{1}{9}$ and common ratio $\frac{1}{3}$, we find:
$$T_{n} < \frac{\frac{1}{9}(1-\frac{1}{{3}^{n}})}{1-\frac{1}{3}}=\frac{1}{6}-\frac{1}{6}\times \frac{1}{{3}^{n}} < \frac{1}{6}.$$

Therefore, we conclude that $T_{n}＜\frac{1}{6}$, and we encapsulate this final answer as $\boxed{T_{n}＜\frac{1}{6}}$.