Contradiction while using the convolution sum for a non-LTI system
You've some troubles about the LTI property and the impulse response. First of all let me give you the answer to your quiz question.

Given two LTI systems $\mathcal{T}_1$ and $\mathcal{T}_2$ with impulse responses :

$$h_1[n] = \delta[n+3]$$
$$h_2[n] = \delta[5-n] = \delta[n-5]$$

you can compute the output of the cascade $$y[n] = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x[n] \} \} $$ for the input signal $x[n] = n$ as :

$$ y_1[n] = \mathcal{T}_1 \{ x[n] \} \implies y_1[n] = n \star \delta[n+3] = (n+3) $$

and 

$$ y_2[n] = \mathcal{T}_2 \{ y_1[n] \} \implies y_2[n] = (n+3) \star \delta[n-5] = (n-2) $$

so your output is $$y[n] = n-2$$

Note that $h_2[n] = \delta[5-n]$ and $h_2[n] = \delta[n-5]$ are the same systems, and yield the same output. 

Note again that both systems are declared to be LTI and are given their respective impulse responses $h_1[n]$ and $h_2[n]$ , therefore their cascade will also be LTI and furthermore the output does not depend on the ordering of the systems; i.e.

$$y[n] = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x[n] \} \} = \mathcal{T}_1 \{ \mathcal{T}_2 \{ x[n] \} \}  $$

Now coming to your own questions:

1- You cannot use convolution sum to deduce LTI property. You cannot form a convolution sum if the system is not already LTI.

2- The question seems ok:

Note: your claim that $ h[n] = \delta[5-n] $ , signifies a non-LTI system is wrong...