Rate of a Chemical Reaction

$N_2+3H_2\rightleftharpoons2NH_3$ Rate of reaction $=-\frac{d[N_2]}{dt}=-\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{1}{2}\frac{d[NH_3]}{dt}$ So $-\frac{d[N_2]}{dt}=\frac{1}{2}\frac{d[NH_3]}{dt}$ is correct.

For $2N_2O_5 \rightarrow 4NO_2 + O_2$: Rate $= -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Substituting the three given expressions: $\frac{1}{2}k[N_2O_5] = \frac{1}{4}k'[N_2O_5] = k''[N_2O_5]$
$\frac{k}{2} = \frac{k'}{4} = k''$
From $\frac{k}{2} = \frac{k'}{4}$: $k' = 2k$
From $\frac{k}{2} = k''$: $k'' = \frac{k}{2}$
So k′ = 2k and k″ = k/2.

For $N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = 2\frac{d[O_2]}{dt}$
$\frac{d[NO_2]}{dt} = 2\times\left(-\frac{d[N_2O_5]}{dt}\right) = 2\times6.25\times10^{-3}$
$\frac{d[NO_2]}{dt} = 1.25\times10^{-2}$ mol $L^{-1}s^{-1}$
$\frac{d[O_2]}{dt} = \frac{1}{2}\times\left(-\frac{d[N_2O_5]}{dt}\right) = \frac{1}{2}\times6.25\times10^{-3}$
$\frac{d[O_2]}{dt} = 3.125\times10^{-3}$ mol $L^{-1}s^{-1}$

For $N_2 + 3H_2 \rightarrow 2NH_3$: Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
$-\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2}\frac{d[NH_3]}{dt} = \frac{3}{2}\times2\times10^{-4}$
$-\frac{d[H_2]}{dt} = 3\times10^{-4}$ mol $L^{-1}s^{-1}$

For the given reaction: Rate $= -\frac{1}{5}\frac{d[Br^-]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt}$
The negative sign appears with $Br^-$ because its concentration decreases with time.
$\frac{1}{3}\frac{d[Br_2]}{dt} = -\frac{1}{5}\frac{d[Br^-]}{dt}$
$\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}$

For the reaction $2A + B \rightarrow 3C + D$, the rate is expressed as:
Rate $= -\dfrac{1}{2}\dfrac{d[A]}{dt} = -\dfrac{d[B]}{dt} = +\dfrac{1}{3}\dfrac{d[C]}{dt} = +\dfrac{d[D]}{dt}$
The concentration of a reactant decreases with time, so a negative sign is used to make the rate positive.
The concentration of a product increases with time, so a positive sign is used.
C is a product, so the correct expression for it is $+\dfrac{d[C]}{3\,dt}$.
$-\dfrac{d[C]}{3\,dt}$ would be a negative quantity, so it does not express the reaction rate.

For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, the rate is:
Rate $= -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = +\dfrac{1}{2}\dfrac{d[NH_3]}{dt}$
Taking the last two terms: $\dfrac{1}{2}\dfrac{d[NH_3]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt}$
Multiplying both sides by 2:
$\dfrac{d[NH_3]}{dt} = -\dfrac{2}{3}\dfrac{d[H_2]}{dt}$
This agrees with the stoichiometry: 3 moles of $H_2$ consumed give 2 moles of $NH_3$.