Obtain the steady-state solution of the problem $$\frac{\partial^2u}{\partial x^2}+\gamma^2(u-T)=\frac{1}{k}\frac{\partial u}{\partial t}, \ 0<x<a, \ t>0,$$ $$u(0,t)=T, \ u(a,t)=T, \ 0<t,$$ $$u(x,0)=T_1\frac{x}{a}, \ 0<x<a.$$
Since steady-state doesn't depend on time then we will have: $$\frac{\partial^2v}{\partial x^2}+\gamma^2(v-T)=0, \ 0<x<a,$$ $$v(0)=T, \ v(a)=T, $$
but the solution given is $v(x)=T$ which I don't know how they obtained?
I believe that the equation should read $\frac{d^2 v}{dx^2} - \gamma^2 (v-T) = 0$.
Use the substitution $V \equiv v - T$. Note that the derivatives of $V$ are that same as that of $v$ because the constant term $T$ drops upon differentiation. The differential equation reduces to
\begin{align*} \frac{d^2 V}{dx^2} - \gamma^2 V & = 0 \\ V(0) = 0, V(a) & = 0 \end{align*}
which has solution $V = A \exp(\gamma x)+ B \exp(-\gamma x)$
We have $V(0) = A + B = 0$ and $V(a) = A \exp(\gamma a) + B \exp(- \gamma a) = 0$, hence $A\left[ \exp(\gamma a) - \exp(-\gamma a) \right] = 0$. Note that $\exp(\gamma a) - \exp(-\gamma a) = 0$ if and only if $\gamma a = 0$ which is not the case and therefore $A = 0$.
We may conclude that $V(x) = 0$ hence $v(x) = T$.
In case the equation really reads $\frac{d^2 v}{dx^2} + \gamma^2 (v-T) = 0$, you get the general solution
\begin{align*} V(x) = A \cos(\gamma x) + B \sin(\gamma x) \end{align*}
In this case you will find $v(x) = T$ under the assumption that $\gamma a \neq n \pi$, $n \in \mathbb{N}$.