My strategy was to prove that the given chi squared distribution converges to a normal distribution and then approximate the probability as a Normal distribution, not sure if that's the correct way to do it. I am so confused with this question, would appreciate the help!
I need to find P(1.75 < X < 2.5) but I am confused as to how I do it here.
The description of $\ \bar{X}\ $ in the title of your question isn't completely clear, but the only sense I can make of it is that $$ \bar{X}=\frac{1}{36}\sum_{j=1}^{36}Z_i $$ where $\ Z_1,Z_2,\dots,Z_{36}\ $ are independent random variables with $\ Z\sim\chi_2^2\ $—that is each $\ Z_i\ $ follows a chi-squared distribution with $2$ degrees of freedom. Now the chi-squared distribution with $2$ degrees of freedom is just that of the sum of the squares of two independent standard normal variates. It follows that $\ \sum_\limits{j=1}^{36}Z_i\ $ must have the same distribution as $$ \sum_{i=1}^{72}Y_i^2\ , $$ where $\ Y_i\ $ are independent standard normal variates—that is, a chi-squared distribution with $72$ degrees of freedom. That number of degrees of freedom is sufficiently large, however, that the distribution should be reasonably well approximated by a normal distribution with mean $\ 72\ $ and standard deviation $\ 12\ $, so your strategy of using that approximation should be acceptable.
The exact probability is given by \begin{align} P(1.75<\bar{X}<2.75)&=P\left(36\times1.75<\sum_{j=1}^{36}Z_i<36\times2.75\right)\\ &=\chi_{72}^2\left(99\right)-\chi_{72}^2\left(63\right)\ , \end{align} which you can evaluate either by looking up the values of $\ \chi_{72}^2\left(99\right)\ $ and $\ \chi_{72}^2\left(63\right)\ $ in a table of the chi-squared distributions, or plugging the arguments $\ 99\ $ and $\ 63\ $ into a chi-squared calculator, such as the one provided by WolframAlfa, which gives $\ \chi_{72}^2\left(99\right)\approx0.980857\ $ and $\ \chi_{72}^2\left(63\right)\approx0.233443\ $. The normal approximation, on the other hand, gives $\ \mathcal{N}_{(72,12)}(99)\approx0.987776\ $ and $\ \mathcal{N}_{(72,12)}(63)\approx0.226627\ $, resulting in an approximation which is about $\ 1.8\%\ $ too high.
Addendum
I've just realised that you can use WolframAlfa to get the values of $\ \chi_{72}^2\left(99\right)-\chi_{72}^2\left(63\right)\ $ and $\ \mathcal{N}_{(72,12)}(99)-$$\,\mathcal{N}_{(72,12)}(63)\ $ directly, without calculating the upper and lower bounds separately. The results to $6$ decimal places are $\ 0.747414\ $ for the first quantity, and $\ 0.761148\ $ for the second.