Step 1:

Let $X$ be the random variable denoting the life of a pair of army shoes.

$X\sim N(8,2^2)$

Step 2:

A pair of shoes would require replacement within 12 months if its life is less than 12 months.

$\therefore$ we have to find $P(X < 12)$

Step 3:

Let $Z$ be the standard normal variate.

$Z=\large\frac{X-\mu}{\sigma}$

$\;\;\;=\large\frac{X-8}{2}$

When $X=12$

$\;\;\;=\large\frac{12-8}{2}$

$\;\;\;=\large\frac{4}{2}$

$\;\;\;=2$

Step 4:

$\therefore P(X < 12)=P(Z < 2)$

$\qquad\qquad\;\;\;\;=0.5+P(0 < Z < 2)$

$\qquad\qquad\;\;\;\;=0.5+0.4772$

$\qquad\qquad\;\;\;\;=0.9772$

Step 5:

Out of 5000 pairs (N=5000) the number of pairs that are expected to be replaced within 12 months=$5000\times 0.9772$

$\Rightarrow 4886$