Find the last digit

Consider the first term: $31^{389}$

$1$ raised to any number is $1$ . So the units digit of the first term is $1$ .

Consider the second term: $5(48^{521})$

$8^1=8, 8^2=64,8^3=512,8^4=4096,8^5=32768$ . The units digit repeat in every cycle of $4$ . The units digit of $48^{521}$ is:

$\dfrac{521}{4}=130$ remainder $1$ . Since $1$ is the remainder, the units digit is $8$ . So the units digit of the second term $(5\cdot 8=40)$ is $0$ .

Consider the third term: $2^{85} \cdot 3^{92}$

$2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32$ . The units digit repeat in every cycle of $4$ . The units digit of $2^{85}$ is:

$\dfrac{85}{4}=21$ remainder $1$ . Since $1$ is the remainder, the units digit is $2$ .

$3^1=3,3^2=9,3^3=27,3^4=81,3^5=243$ . The units digit repeat in every cycle of $4$ . The units digit of $3^{92}$ is:

$\dfrac{92}{4}=23$ . Since there is no remainder, the units digit is $1$ . The units digit of the third term is $2\cdot 1=2$ .

Consider the fourth term: $13^{1095}$

$3^1=3,3^2=9,3^3=27,3^4=81,3^5=243$ . The units digit repeat in every cycle of $4$ . The units digit of $13^{1095}$ is:

$\dfrac{1095}{4}=273$ remainder $3$ . Since $3$ is the remainder, the units digit is $7$ .

Hence,

$1+0-2+7=\boxed{6}$