The key fact to hold onto here is that a standard normal curve and a $t_1$ (Cauchy) curve are both symmetric, single-peaked bell shapes centered at $0$, but they do not have the same "spread". The normal curve is tall and thin near the centre, while the Cauchy curve is shorter at the peak but drags out into much heavier tails.
Peak heights: $g(0) = \frac{1}{\sqrt{2\pi}} \approx 0.399$ against $h(0) = \frac{1}{\pi} \approx 0.318$. The normal peak is clearly taller, so $g(0) \ne h(0)$, statement (D) fails right away.
Because both curves are mirror images of themselves about $0$, exactly half of each distribution's probability sits on the left of $0$ and half on the right, no matter how the shapes differ elsewhere. That gives $G(0) = H(0) = 0.5$ automatically, so statement (A) is true.
Now picture sliding along the positive $x$-axis. Near $0$, the normal curve sits above the Cauchy curve since it is taller there. But because the Cauchy tail decays only like $1/y^2$ while the normal tail decays like $e^{-y^2/2}$, far away from $0$ the Cauchy curve ends up above the normal curve. The two curves must cross exactly once for positive $y$: that crossing point is $c$, where $g(c) = h(c)$.
So on the strip from $0$ to $c$, the normal curve is on top, meaning the normal distribution picks up more probability in that strip than the Cauchy does. That extra probability pushes $G(c)$ above $H(c)$, i.e. $G(c) > H(c)$, the opposite of what statement (B) claims, so (B) is false.
Mirror that strip to the negative side using symmetry. If the normal has more mass than the Cauchy in $(0,c)$, then by the mirror image, the normal has less mass than the Cauchy in $(-c,0)$, since both distributions must still total to 1 on each side. Less mass to the left of $-c$ under the normal than under the Cauchy means $G(-c) < H(-c)$, so statement (C) holds.
The correct statements are (A) and (C).
\[ \boxed{\text{(A) and (C)}} \]