1. Write the given reactions

$\text{(A)}\quad 2\,CO(g)+O_2(g) \to 2\,CO_2(g), \qquad \Delta H_1^\circ=-x$

$\text{(B)}\quad C(\text{graphite})+O_2(g) \to CO_2(g), \qquad \Delta H_2^\circ=-y$

We need to find $\Delta H^\circ$ for:

$C(\text{graphite})+\frac{1}{2}O_2(g) \to CO(g)$

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2. First simplify reaction (A)

Divide reaction (A) by 2:

$CO(g)+\frac{1}{2}O_2(g) \to CO_2(g)$

So its enthalpy becomes:

$\Delta H^\circ = -\frac{x}{2}$

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3. Use Hess's law

We know:

$C(\text{graphite})+O_2(g) \to CO_2(g) \qquad \Delta H^\circ=-y$

Also:

$CO(g)+\frac{1}{2}O_2(g) \to CO_2(g) \qquad \Delta H^\circ=-\frac{x}{2}$

Reverse the second equation:

$CO_2(g) \to CO(g)+\frac{1}{2}O_2(g)$

Then:

$\Delta H^\circ=+\frac{x}{2}$

Now add it to reaction (B):

$C(\text{graphite})+O_2(g) \to CO_2(g) \qquad (-y)$

$CO_2(g) \to CO(g)+\frac{1}{2}O_2(g) \qquad \left(+\frac{x}{2}\right)$

Adding,

$C(\text{graphite})+O_2(g) \to CO(g)+\frac{1}{2}O_2(g)$

Cancel $\frac{1}{2}O_2$ from both sides:

$C(\text{graphite})+\frac{1}{2}O_2(g) \to CO(g)$

Hence,

$\Delta H^\circ=-y+\frac{x}{2}$

$\Delta H^\circ=\frac{x-2y}{2}$

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4. Match with the options

$\boxed{\Delta H^\circ=\frac{x-2y}{2}}$

So the correct option is:

$\boxed{\text{A}}$