How to deal with Z-score greater than 3? In a standard normal distribution how do I deal with a $Z$ value greater than 3? 
I know that z-score ranges form -3 to 3   
Consider this one ...   
mean = 70, standard deviation = 4  
I need to find $P(65 < X < 85)$.
Transforming to standard normal gives $P(-1.25 < Z < 3.75)$
How to deal with the $3.75$?  
Edit *Actually it's not greater than 3, $z < 3.75$. I meant smaller than, sorry. Should I assume it's just 0.4990 or what?
 A: The standard normal ranges from $-\infty$ to $\infty$.
Your problem appears to be that your table doesn't go further. 
Your question should therefore be modified to ask "*How do I deal with the fact that my table doesn't go as high as my $Z$ value?*"
[Note that in your last paragraph, you have become confused. The region you're evaluating probability for is $Z<3.75$ but the boundary value of $Z$ you're trying to look up in the table (the $3.75$) is $>3$, as in your title.]
It seems like not having the value in your table would be a problem, but it's a very small one $-$ since your answer for $P(0<Z<3.75)$ can't be smaller than $P(0<Z<3)\approx 0.4999$ and can't be larger than $P(0<Z<\infty)=0.5$, you shouldn't have much difficulty narrowing the answer down to 3 significant figures of accuracy even so.
Additional accuracy (though I really don't think you need it) can be obtained by many methods. Here are three:
i) finding better tables (these seem to be of the same form as the ones you're apparently using)
ii) using a package that will evaluate standard normal cdfs for you. I just used R (simply typing pnorm(3.75) to obtain $P(-\infty<Z<3.75)$).
iii) using numerical integration to approximate the area between 3 and 3.75. For example, via Simpson's rule, a single interval (3 points) gives 0.0017 (the correct answer is 0.0013 to 4dp). Alternatively, because the density is convex in this region (indeed, as whuber points out in comments, convex for $Z>1$ and $Z< -1$), the integral will be bounded below by the midpoint rule and above by the trapezoidal rule, which usefully bounds where the answer can lie
But, really, just using the limits provided by 3 and $\infty$ is plenty, I imagine.
A: For $z >0$, the right tail of the standard normal distribution 
(that is, the area to the right of $z$) which is often denoted by 
$Q(z)$ is bounded as follows:
$$\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z} - \frac{1}{z^3}\right )
\ < \ Q(z) \ < \ \frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right ).$$
See, for example this answer on math.SE for a proof. The bounds blow up/down to $\pm \infty$ as $z \to 0$ but are quite useful in the
regions not covered by typical tables of the cumulative distribution function
of the standard normal random variable. For example,
$$0.0000873 < Q(3.75) < 0.0000940$$ while the actual value is slightly larger than $0.0000884$
A: Let me repeat (and correct) what I've said in my comment ad reply to your edit.
You have to transfer from $X$ to $Z$ in order to use a z-score table. Since a z-score table contains a small finite subset of values, you often must settle for an approximation. So you could also settle for $P(Z<-3)\approx 0$ and $P(Z< 3)\approx 1$ (NB: $P(Z>3)\approx 1$ was a typo, sorry.)
As to $P(-1.25<Z<3.75)$, I'll use this z-score table:
$$P(-1.25<Z<3.75)=P(Z<3.75)-P(Z<-1.25)\approx 1-0.1056=0.8944$$
A: The z (i.e., normal) distribution is not bounded. $\mathcal N(\mu=70,\sigma=4)$ is not standard normal either – that refers to $\mathcal N(0,1)$. If you're wondering what the p value is for z = 3.75, you can find it in r with pnorm(3.75). (You could also use pnorm(85,70,4).) The result is p =  0.9999116.
If you want an exact p value, I think you're going to have an easier time getting it in R or some other statistical software than by dealing with the quantile function directly...but FWIW, here's that equation:

$$F^{-1}(p)
      = \mu + \sigma\Phi^{-1}(p)
      = \mu + \sigma\sqrt2\,\operatorname{erf}^{-1}(2p - 1), \quad p\in(0,1)$$

In the above, $\rm erf$ refers to the error function.
In light of your comments and edit to the question, I think I should decline to provide more than this as a "hint" in following our policy on self-study questions. 
