Positive at both times

Idea. Write $X=W_1$ and $Y=W_2-W_1$, two independent standard normals. The point $(X,Y)$ has a uniformly distributed direction, and the conditions $X>0$ and $X+Y>0$ cut out a sector. The answer is the sector's angle divided by $2\pi$.

Step 1. The conditions are $X>0$ and $X+Y>0$.

Step 2: the sector. $X>0$ is the right half-plane, angles between $-90°$ and $90°$. $X+Y>0$ is the half-plane above the line $Y=-X$, angles between $-45°$ and $135°$. Both hold for angles between $-45°$ and $90°$: a sector of $135°$.

Step 3. The standard normal distribution in the plane is rotation invariant, so $$\mathbb{P}(W_1>0,W_2>0)=\frac{135}{360}=\frac{3}{8}.$$

Remarks. Given $W_1>0$, the probability that $W_2>0$ is $\frac{3/8}{1/2}=\frac{3}{4}$. In general, two standard normals with correlation $\rho$ are both positive with probability $\frac{1}{4}+\frac{\arcsin \rho }{2\pi }$; here $\rho =\mathrm{corr}(W_1,W_2)=\frac{1}{\sqrt{2}}$, so $\arcsin \rho =\frac{\pi }{4}$ and the formula gives $\frac{3}{8}$ again.