11. Let the quadratic function $f(x)$ attain its maximum value of 5 when $x=a, a>0$. The quadratic function $g(x)$ has a minimum value of -2, and $g(a)=25, f(x)+g(x)=x^{2}+16 x+13$.
(1) Find the value of $a$; (2) Find the expression for $g(x)$.
11. (1) According to the problem, let $f(x)=k(x-a)^{2}+5$, at this time $g(x)=x^{2}+16 x+13-k(x-a)^{2}-5$. Also, $g(a)=25$, then $a^{2}+16 a+13-5=25, a>0$, we get $a=1$.
(2) $a=1$, then $g(x)=x^{2}+16 x+13-k(x-1)^{2}-5=(1-k) x^{2}+2(k+8) x+8-k$, $k<0$. By $g_{\text {min }}=-2$, then $8-k-\frac{(k+8)^{2}}{1-k}=-2$, we get $k=-2$. Thus, $g(x)=3 x^{2}+12 x+10$.