[Solution] (1) Let the area of low-priced houses form a series {an}, from which we can know that {an} is arithmetic progression.

Where a 1 = 250 and d = 50, then Sn=250n+=25n2+225n,

Let 25n2+225n≥4750, that is, n2+9n- 190≥0, where n is a positive integer and ∴n≥ 10.

By the end of 20 13, the accumulated low-rent housing area in this city over the years is not less than 47.5 million square meters for the first time.

(2) Let the newly-built housing area form a series {bn}, which means that {bn} is a geometric series.

Where B 1 = 400 and Q = 1.08, bn = 400 (1.08) n-1.85.

An & gt0.85 bn, in which 250+(n-1) 50 > 400 (1.08) n-10.85.

The minimum positive integer n=6 satisfying the above inequality is obtained by the counter.

By the end of 2009, the area of low-priced housing built in that year accounted for more than 85% of the housing built in that year for the first time.