Let $t>0$ be a given constant. In each of the following cases, find (with justification) the value of $x$ that maximizes the probability $P(x<X<x+t)$
With $X$ ~ $N(\mu,\sigma^2)$
Trying to maximize the probability I set $P(x<X<x+t)=1$
From there I have $\Phi(\frac{x+t-\mu}{\sigma})-\Phi(\frac{x-\mu}{\sigma})=1$
I'm find that x has to go to $-\infty$ and (x+t) to $+\infty$
I made this little graph just to explain my point.
But I can't think of any easy way to proceed. Any hint would be greatly appreciated..
You can write
$$g(x) = \mathbb P (x < X < x+t) = \int_x^{x+t}\, f_X(s)\,ds$$
where $f_X$ is the pdf of $X$. Then you need to maximize $g(x)$, and this is a single-variable calculus problem.
For your particular case of $X$, though, there's a fairly visual intuition that helps find the answer with much less work. Consider the graph of $f_X$; then $g(x)$ is the area beneath the graph on $[x,x+t]$. As $x$ varies, you move the window $[x,x+t]$.
You need to find the window which yields most area, and at this point I think it's intuitively clear that the window we seek will be the one whose midpoint lies precisely on $\mu$.
Just for kicks, let's validate our intuition. First, observe that
$$f_X(x) = \frac1{\sqrt{2\pi\sigma^2}}\,\exp\left(-\frac{{(x-\mu)}^2}{2\sigma^2}\right),$$
so we have $f_X > 0$, which implies $0< g < 1$. Additionally, $f_X(x) \to 0$ as $x\to \pm \infty$, from which it follows that $g(x) \to 0$ as $x\to \pm \infty$.
For any fixed $c \in \Bbb R$, we can hence find $M>0$ such that whenever $x\not\in (-M, M)$, $g(x) < g(c)$ and also $g(M), g(-M) < g(c)$.
It follows that our maximum exists and must be attained, because the problem boils down to maximization over a compact interval $[-M, M]$. Moreover, since the maximum does not lie on the endpoints, we are guaranteed that the maximum must be a critical point.
By the Fundamental Theorem of Calculus, $g'(x) = f_X(x+t) - f_X(x)$ and hence
$$g'(x) = 0 \iff \exp\left(-\frac{{(x+t-\mu)}^2}{2\sigma^2}\right) = \exp\left(-\frac{{(x-\mu)}^2}{2\sigma^2}\right)$$
Since $\exp$ is injective over the reals, we conclude
\begin{align} g'(x) = 0 &\iff {(x+t-\mu)}^2 = {(x-\mu)}^2 \\&\iff {(x+t-\mu)}^2 - {(x-\mu)}^2 = 0 \\&\iff \Big((x+t-\mu)+(x-\mu)\Big)\cdot \Big((x+t-\mu)-(x-\mu)\Big) = 0 \\&\iff \mu = \frac{x+(x+t)}2 \quad \text{ or } \quad t = 0 \end{align}
Because $t>0$ from our hypothesis, our only option is $\mu = \frac{x+(x+t)}2$, which precisely means that the midpoint of your windows lies on $\mu$. We of course may rewrite as
$$x = \mu - \frac t2.$$
Moreover, this is $g$'s only critical point (on all of the real line), so it must be the maximum.