Answer
a. True. For every object x, you can choose an object y that is not the same as x and has a different color than x.
b. Formal: $\forall x$(Object(x) $\rightarrow$ ($\exists y$ (Object(y) $\land$ $x \neq y$ $\land$ DifferentColor(x,y))))
c. Negation: $\exists x$ (Object(x) $\land$ ($\forall y$ (~Object(y) $\lor$ $x = y$ $\lor$ ~DifferentColor(x,y))))
Work Step by Step
c. Negation:
~($\forall x$(Object(x) $\rightarrow$ ($\exists y$ (Object(y) $\land$ $x \neq y$ $\land$ DifferentColor(x,y)))))
$\equiv$ $\exists x$ ~(Object(x) $\rightarrow$ ($\exists y$ (Object(y) $\land$ $x \neq y$ $\land$ DifferentColor(x,y))))
(by the law for negating a $\forall$ statement)
$\equiv$ $\exists x$ (Object(x) $\land$ ~($\exists y$ (Object(y) $\land$ $x \neq y$ $\land$ DifferentColor(x,y))))
(by the law for negation an if-then statement)
$\equiv$ $\exists x$ (Object(x) $\land$ ($\forall y$ ~(Object(y) $\land$ $x \neq y$ $\land$ DifferentColor(x,y))))
(by the law for negating an $\exists$ statement)
$\equiv$ $\exists x$ (Object(x) $\land$ ($\forall y$ (~Object(y) $\lor$ $x = y$ $\lor$ ~DifferentColor(x,y))))
(by De Morgan's law)
