Ytukyg

As in the attached picture below, how to find the values of I2 and I3?
enter image description here

Per your text $$I = T_{11} + T_{22}+T_{33}\II =frac{T_{ij}T_{ij} - I^2}2\III=det mathbf T$$

You've proven the first formula. The last formula is almost trivial to prove without expanding it in elements. After all, $III$ is the constant term of the characteristic polynomial $p(lambda) = det(mathbf T - lambdamathbf E)$, where $mathbf E$ is the identity matrix. The constant term of $p$ is the value of $p(lambda)$ when $lambda = 0$.
Therefore $$III = p(0) = det(mathbf T - 0cdot mathbf E) = det mathbf T$$

That leaves only $II$. $$begin{align}I^2 &= (T_{11} + T_{22} + T_{33})^2\&=T_{11}^2 + T_{22}^2 + T_{33}^2+2(T_{11}T_{22} + T_{11}T_{33} + T_{22}T_{33})end{align}$$

I had assumed without bothering to check that $T_{ij}T_{ij}$ was making use of the Einstein summation convention. But that isn't true. Your calculation of $II$ is correct:
$$II = T_{12}T_{21} + T_{13}T_{31} + T_{23}T_{32} - T_{11}T_{22} - T_{11}T_{33} - T_{22}T_{33}$$
But if we interpret this per the convention
$$T_{ij}T_{ij} = sum_{i, j} T_{ij}^2 = T_{11}^2 + T_{22}^2 +T_{33}^2 +T_{12}^2 +T_{21}^2 +T_{13}^2 +T_{31}^2 +T_{23}^2 +T_{32}^2$$ the expression clearly does not give the correct result.

It is actually well-known that for 3x3 matrices, $II = frac{operatorname{tr}(T^2) - (operatorname{tr}(T))^2}2 = frac{operatorname{tr}(T^2) - I^2}2$. So apparently $T_{ij}T_{ij}$ is supposed to be $operatorname{tr}(T^2)$, but I see no way that this makes sense as a notational shorthand.

So instead, let me use the notation which does make sense (though is not a particularly good way to calculate the coefficient, since it involves a lot of wasted calculation):

$$mathbf T^2 = begin{bmatrix}T_{11}^2 + T_{12}T_{21} + T_{13}T_{31} & ... & ...\ ... & T_{21}T_{12} + T_{22}^2 + T_{23}T_{32} & ...\...&...&T_{31}T_{13} + T_{32}T_{23} + T_{33}^2end{bmatrix}$$
where I haven't bothered to expand the entries that will not contribute to the trace. So, $$operatorname{tr}(mathbf T^2) = T_{11}^2 + T_{22}^2 +T_{33}^2 + 2(T_{12}T_{21} + T_{13}T_{31} + T_{31}T_{13})$$

It is should be easy now to get from $frac{operatorname{tr}(T^2) - I^2}2$ to your coefficient for $lambda$.