This question is now old enough that I can give you a solution without ruining your homework. As you have pointed out in your question, to compute this probability, you need to find the distribution of $D=X-Y$. One of the properties of the normal distribution is that any linear combination of independent normal random variables is also a normal random variable, so this establishes that $D \sim \mathcal{N}$, and it remains only to find the mean and variance of this random variable. You have correctly derived the mean, but incorrectly derived the variance.
To obtain the mean and variance of $D$ we apply standard rules for the mean and variance of linear functions of random variables. Since the mean is a linear operator and the variance is a quadratic operator, we have:
$$\mathbb{E}(D) = \mathbb{E}(X-Y) = \mathbb{E}(X) - \mathbb{E}(Y) = 1-2 = -1.$$
$$\mathbb{V}(D) = \mathbb{V}(X-Y) = \mathbb{V}(X) + (-1)^2 \mathbb{V}(Y) = 2+3 = 5.$$
Thus, you have the probability:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X>Y) = \mathbb{P}(X-Y>0)
&= \mathbb{P}(D>0) \\[6pt]
&= \mathbb{P} \bigg( \frac{D+1}{\sqrt{5}} > \frac{0+1}{\sqrt{5}} \bigg) \\[6pt]
&= 1-\Phi \bigg( \frac{0+1}{\sqrt{5}} \bigg) \\[6pt]
&= 0.3273604, \\[6pt]
\end{aligned} \end{equation}$$
where $\Phi$ denotes the cumulative distribution function for the standard normal distribution.
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:
$$P(-1.25<Z<3.75)=P(Z<3.75)-P(Z<-1.25)\approx 1-0.1056=0.8944$$
Best Answer
Compute $1 - F(μ + 3σ)$, where $F$ is the cumulative distribution function, $μ$ the mean, and $σ$ the standard deviation. When the distribution is normal, this comes out to .00135.