The value of $$\sin^230^{\circ}\cos^245^{\circ}+4\tan^230^{\circ}+\frac{1}{2}\sin^290^{\circ}+2\cos90^{\circ}$$ is:

$$\sin^230^{\circ}\cos^245^{\circ}+4\tan^230^{\circ}+\frac{1}{2}\sin^290^{\circ}+2\cos90^{\circ}=\left(\frac{1}{2}\right)^2\left(\frac{1}{\sqrt{2}}\right)^2+4\left(\frac{1}{\sqrt{3}}\right)^2+\frac{1}{2}\left(1\right)^2+2\left(0\right)$$

$$=\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)+4\left(\frac{1}{3}\right)+\frac{1}{2}$$

$$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}$$

$$=\frac{3+32+12}{24}$$

$$=\frac{47}{24}$$

Hence, the correct answer is Option B