My first problem!

We note that the integrand of $\displaystyle \int_\frac \pi 4^{\frac {65\pi}4} \frac {dx}{(1+2^{\sin x})(1+2^{\cos x})}$ is periodic with a period of $2\pi$ . The integral has a domain $\dfrac {65\pi}4 - \dfrac \pi 4 = 16\pi$ or 8 cycles. Therefore,

$\begin{aligned} I & = \int_\frac \pi 4^{\frac {65\pi}4} \frac {dx}{(1+2^{\sin x})(1+2^{\cos x})} \\ & = 8 \int_0^{2\pi} \frac {dx}{(1+2^{\sin x})(1+2^{\cos x})} & \small \color{#3D99F6} \text{Using the identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 82 \int_0^{2\pi} \left( \frac 1{(1+2^{\sin x})(1+2^{\cos x})} + \frac 1{(1+2^{\sin (2\pi- x})(1+2^{\cos (2\pi-x)})} \right) dx \\ & = 4 \int_0^{2\pi} \left( \frac 1{(1+2^{\sin x})(1+2^{\cos x})} + \frac 1{(1+2^{-\sin x})(1+2^{\cos x})} \right) dx \\ & = 4 \int_0^{2\pi} \left( \frac 1{(1+2^{\sin x})(1+2^{\cos x})} + \frac {2^{\sin x}}{(1+2^{\sin x})(1+2^{\cos x})} \right) dx \\ & = 4 \int_0^{2\pi} \frac 1{1+2^{\cos x}} dx & \small \color{#3D99F6} \text{Integral is symmetrical about }\pi. \\ & = 8 \int_0^\pi \frac 1{1+2^{\cos x}} dx & \small \color{#3D99F6} \text{Using the identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = 4 \int_0^\pi \left( \frac 1{1+2^{\cos x}} + \frac 1{1+2^{\cos (\pi-x)}} \right) dx \\ & = 4 \int_0^\pi \left( \frac 1{1+2^{\cos x}} + \frac 1{1+2^{-\cos x}} \right) dx \\ & = 4 \int_0^\pi \left( \frac 1{1+2^{\cos x}} + \frac {2^{\cos x}}{1+2^{\cos x}} \right) dx \\ & = 4 \int_0^\pi 1 \ dx \\ & = 4 \pi \end{aligned}$

$\implies k = \boxed{4}$