SG doesn't only stand for Singapore...

The given product can be written as:

$\displaystyle\prod_{r=0}^{4}\frac{(10+12r)^{4}+4(3)^{4}}{(4+12r)^{4}+4(3)^{4}}$

Now we shall use the Sophie Germain Identity

$a^{4}+4b^{4}=(a^{2}+2b^{2}+2ab)(a^{2}+2b^{2}-2ab)$

to obtain:

$\displaystyle\prod_{r=0}^{4}\frac{\bigg( (10+12r)^{2}+2(3)^{2}+2\times 3\times (10+2r)\bigg)\bigg( (10+12r)^{2}+2(3)^{2}-2\times 3\times (10+2r)\bigg)}{\bigg( (4+12r)^{2}+2(3)^{2}+2\times 3\times (4+2r)\bigg)\bigg( (4+12r)^{2}+2(3)^{2}-2\times 3\times (4+2r)\bigg)}$

$=\displaystyle\prod_{r=0}^{4}\frac{(144r^{2}+312r+178)(144r^{2}+168r+58)}{(144r^{2}+168r+58)(144r^{2}+24r+10)}$

$=\displaystyle\prod_{r=0}^{4}\frac{(12(r+1)+1)^{2}+9} {(12r+1)^{2}+9}$

$=\dfrac{61^{2}+9}{1^{2}+9}$

$=\boxed{373}$