Calculus is very easy

$\lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 1+\cos { 2x } }{ { (\pi -2x) }^{ 2 } } } \\ Since\quad it\quad is\quad a\quad \frac { 0 }{ 0 } \quad form,\quad so\quad we\quad need\quad to\quad simplify\quad it.\\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 1-\cos { (\pi -2x) } }{ { (\pi -2x) }^{ 2 } } } \\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 1-(1-2\sin ^{ 2 }{ \frac { (\pi -2x) }{ 2 } } ) }{ { (\pi -2x) }^{ 2 } } } \quad \quad \quad \quad \quad \quad \{ Using\quad \cos { x } =\quad 1-2\sin ^{ 2 }{ \frac { x }{ 2 } } )\\ Opening\quad the\quad brackets,\\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 2\sin ^{ 2 }{ \frac { (\pi -2x) }{ 2 } } }{ { (\pi -2x) }^{ 2 } } } \\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { \sin { { (\frac { (\pi -2x) }{ 2 } })^{ 2 } } }{ { \frac { { (\pi -2x) }^{ 2 } }{ 2 } } } } \\ Multiplying\quad and\quad dividing\quad 2\quad in\quad the\quad denominator,\\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { \sin { { (\frac { (\pi -2x) }{ 2 } })^{ 2 } } }{ { { (\frac { (\pi -2x) }{ 2 } ) }^{ 2 }\quad }2 } } \\ Separating\quad terms,\\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { \sin { { (\frac { (\pi -2x) }{ 2 } })^{ 2 } } }{ { { (\frac { (\pi -2x) }{ 2 } ) }^{ 2 }\quad } } } \quad \times \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } \\ But\quad if\quad x\rightarrow \frac { \pi }{ 2 } ,\quad So,\quad 2x-\pi \rightarrow 0,\quad Or,\quad \pi -2x\rightarrow 0\\ Changing\quad limits\\ \Rightarrow \quad \lim _{ \pi -2x\rightarrow 0 }{ \frac { \sin { { (\frac { (\pi -2x) }{ 2 } })^{ 2 } } }{ { { (\frac { (\pi -2x) }{ 2 } ) }^{ 2 }\quad } } } \times \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } \\ But\quad \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } } =1,\quad So,\quad first\quad term\quad =\quad 1\\ Therefore,\\ \Rightarrow \quad \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } \quad =\quad \frac { 1 }{ 2 } \quad =\quad 0.5$

Substituting $t=\frac{\pi}{2}-x$ we have $\lim_{t\to 0}\frac{1+\cos(\pi-2t)}{4t^2}=\lim_{t\to 0}\frac{1-\cos(2t)}{4t^2}=\lim_{t\to 0}\frac{2\sin^2(t)}{4t^2}=\boxed{0.5}$

Applying Taylor series of $\cos x$ around $x = \pi$ , then $\cos x = -1 + \frac{(x - \pi)^2}{2!} - ....$ $\lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(\pi - 2x)^2} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{(2x - \pi)^2}{2!}}{(\pi - 2x)^2} = 0.5$