Answer
a. False. Take for instance object a. a $\neq$ c, but a and c have the same color.
b. Formal: ∀x(Object(x) → (∃y (Object(y) ∧ (x$\neq$y $\rightarrow$ DifferentColor(x,y))))
c. Negation: $\exists$x (Object(x) $\land$ ($\forall$y ~(Object(y) $\lor$ (x$\neq$y $\land$ ~DifferentColor(x,y)))))
Work Step by Step
c. ~(∀x(Object(x) → (∃y (Object(y) ∧ (x$\neq$y $\rightarrow$ DifferentColor(x,y)))))
$\equiv$ $\exists$x ~(Object(x) → (∃y (Object(y) ∧ (x$\neq$y $\rightarrow$ DifferentColor(x,y)))))
(by the law of negating a $\forall$ statement)
$\equiv$ $\exists$x (Object(x) $\land$ ~(∃y (Object(y) ∧ (x$\neq$y $\rightarrow$ DifferentColor(x,y)))))
(by the law of negating an if-then statement)
$\equiv$ $\exists$x (Object(x) $\land$ ($\forall$y ~(Object(y) ∧ (x$\neq$y $\rightarrow$ DifferentColor(x,y)))))
(by the law of negating an $\exists$ statement)
$\equiv$ $\exists$x (Object(x) $\land$ ($\forall$y ~(Object(y) $\lor$ ~(x$\neq$y $\rightarrow$ DifferentColor(x,y)))))
(by De Morgan's law)
$\equiv$ $\exists$x (Object(x) $\land$ ($\forall$y ~(Object(y) $\lor$ (x$\neq$y $\land$ ~DifferentColor(x,y)))))
(by the law of negating an if-then statement)
