The quantile or probit function, as you can see from the link (see "Computatuon"), is computed with inverse gaussian error function which I hope is downloadable for calculators like TI-89. Look here for instance.

The answer is No, not exactly anyhow.

If you have two quartiles of a normal population then you can find $\mu$ and $\sigma.$ For example the lower and upper quantiles of $\mathsf{Norm}(\mu = 100,\, \sigma = 10)$ are $93.255$ and $106.745,$ respectively.

qnorm(c(.25, .75), 100, 10)
 [1]  93.2551 106.7449

Then $P\left(\frac{X-\mu}{\sigma} < -0.6745\right) = 0.25$ and $P\left(\frac{X-\mu}{\sigma} < 0.6745\right) = 0.75$ provide two equations that can be solved to find $\mu$ and $\sigma.$

qnorm(c(.25,.75))
[1] -0.6744898  0.6744898

However, sample quartiles are not population quartiles. There is not enough information in any normal sample precisely to determine $\mu$ and $\sigma.$

And you are not really sure your sample is from a normal population. If the population has mean $\mu$ and median $\eta,$ then the sample mean and median, respectively, are estimates of these two parameters. If the population is symmetrical, then $\mu = \eta,$ but you say the sample mean and median do not agree. So you cannot be sure the population is symmetrical, much less normal.

Look at a table of the standard normal distribution.

At what values of z it is $\Phi(z)=0.25, \Phi(z)=0.50$ or $\Phi(z)=0.75$ ?

The values of z are the quartiles $Q_1,Q_2$ and $Q_3$.

"Normal distribution is a bell-shaped curve that represents the distribution of a set of data which is symmetrical and centered around the mean.  Key Points In a normal distribution, the mean, median, and mode are all equal. Quartile deviation is a measure of dispersion that indicates the spread of a distribution relative to the interquartile range (IQR). In a normal distribution, the quartile deviation is equal to two-thirds of the standard deviation. To calculate the quartile deviation of a normal distribution, we need to know the standard deviation of the distribution. From the given information, we know that the normal distribution has a mean of 0 and a standard deviation of 1 (since it is a standard normal distribution). The quartile deviation is calculated as follows: Quartile deviation = (Q3 - Q1) 2 where Q3 and Q1 are the upper and lower quartiles, respectively. Using the properties of the normal distribution, we can find the z-scores that correspond to the upper and lower quartiles: Q3 = 0.6745 (since 75% of the data falls below this z-score) Q1 = -0.6745 (since 25% of the data falls below this z-score) Substituting these values into the quartile deviation formula, we get: Quartile deviation = (0.6745 - (-0.6745)) 2 = 1.349 = 23 (rounded to the nearest hundredth). Therefore, the quartile deviation of the normal distribution is 23."

With two quartiles (or any other two percentiles) of a normal distribution you can always compute mean and standard deviation. We start with the formulas to compute quartiles from distribution parameters:

$$Q_1=\mu+CDF^{-1}(0.25)·\sigma=\mu-0.67449·\sigma$$ $$Q_3=\mu+CDF^{-1}(0.75)·\sigma=\mu+0.67449·\sigma$$

Where $CDF^{-1}$ stands for inverse of cumulative distribution function of standard normal distribution, $\mu$ stands for mean and $\sigma$ stands for standard deviation.

Solving the equations for $\mu$ and $\sigma$:

$$\mu=\frac{Q_1+Q_3}{2}$$ $$\sigma=\frac{Q_3-Q_1}{CDF^{-1}(0.75)-CDF^{-1}(0.25)}=\frac{Q_3-Q_1}{2·CDF^{-1}(0.75)}=\frac{Q_3-Q_1}{1.34898}$$

You may be confusing population quantiles with the sample quantiles that estimate them. Your population quantiles are appropriately represented in your figures.

Population quantiles. If random variable $X \sim \mathsf{Norm}(\mu = 100, \sigma = 15),$ then quantiles $.01, .05, .25, .50, .95, .99$ of the distribution can be found in R by using the quantile function qnorm. (The quantile function is sometimes called the 'inverse CDF` function.)

q = round(qnorm(c(.01,.05,.25,.50,.75,.95,.99), 100, 15),3);  q
[1]  65.105  75.327  89.883 100.000 110.117 124.673 134.895 

These quantiles (at vertical lines) can be displayed along with the density function of $\mathsf{Norm}(100, 15)$ as shown in the graph below.

 curve(dnorm(x, 100, 15), 50, 150, col="blue", lwd=2, ylab="PDF",
      main="Density of NORM(100, 15) with Various Quantiles")
   abline(h=0, col="green2");  abline(v=0, col="green2")
   abline(v=q, col="red", lty="dotted", lwd=2)

The total area (representing probability) under the density curve is $1.$ Areas to the left of the three left-most vertical lines are $.01,.05,$ and $.25,$ respectively.

Sample quantiles. If I have a sufficiently large sample from this distribution, then I can find the quantiles of the sample. For example, the 50th sample percentile (quantile .5) is the sample median. These sample quantiles estimate the corresponding population percentiles. Generally speaking, larger samples give better estimates. I will use $n = 1000$ in my example.

set.seed(2020) # for reproducibility
x = round(rnorm(1000, 100, 15), 3)

Here are some summary statistics of the sample, including the sample first quartile (quantile .25), the sample median, and the sample third quartile (quantile .75). The boxplot uses the quartiles [upper and lower edges of the box]and the median [center line inside box], so we show it also.

summary(x)
    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   52.51   89.30   99.14   99.60  109.58  155.54 

boxplot(x, col="skyblue2", horizontal=T,
        main="n=1000; Boxplot of Sample from NORM(100,15)")

Without extra arguments, the R procedure quantile shows the maximum and minimum values in the sample and the three quantiles shown in the summary.

quantile(x)
       0%       25%       50%       75%      100% 
 52.50800  89.30475  99.13750 109.57850 155.54300 

In order to get our full list of quantiles, we need to specify them individually.

samp.q = quantile(x, c(.01,.05,.25,.50,.75,.95,.99));  samp.q
       1%        5%       25%       50%       75%       95%       99% 
 63.76255  74.46450  89.30475  99.13750 109.57850 126.38775 136.60263 

In particular, notice that population quantile .05 (which is $75.327$ from earlier) is estimated by the sample quantile .05 (which is $74.465$ just above).

Finally, we show a histogram of the $n=1000$ observations along with the population density curve. Now the vertical dotted lines show the positions of our chosen sample quantiles.

hist(x, prob=T, col="skyblue2", main="Histogram of Sample")
 curve(dnorm(x, 100, 15), add=T, col="blue", lwd=2)
 abline(v=samp.q, col="purple", lty="dotted", lwd=2)

Numbers of observations at or to the left of the three left-most vertical lines are $10, 50,$ and $250,$ respectively, out of $1000.$

Note: All of the above is about quantiles for a normal distribution because your question deals only with normal distributions. But @Nick Cox makes a good point that quantiles are used similarly for other distributions. For example, here is a plot of an exponential distribution that has rate $\lambda = 0.1$ (hence mean $\mu = 10),$ with vertical lines at the same quantiles used above for the normal distribution.

q = round(qexp(c(.01,.05,.25,.50,.75,.95,.99), 0.1),3);  q
[1]  0.101  0.513  2.877  6.931 13.863 29.957 46.052

curve(dexp(x, 0.1), 0, 60, col="blue", lwd=2, ylab="PDF", n=10001,
      main="Density of EXP(mean=10) with Various Quantiles")
  abline(h=0, col="green2");  abline(v=0, col="green2")
  abline(v=q, col="red", lty="dotted", lwd=2)